An AC source operating at 60. Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and a capacitor ( C = 2.5 μ F). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the capacitor? (c) When the current is zero, what are the magnitudes of the potential difference across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant? (d) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant?

Question

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Answer 1

Textbook Question

Chapter 21, Problem 30P

An AC source operating at 60. Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and a capacitor (C = 2.5 μF). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the capacitor? (c) When the current is zero, what are the magnitudes of the potential difference across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant? (d) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant?

(a)

Expert Solution

To determine

The maximum value of current in the circuit.

Answer to Problem 30P

The maximum current in the circuit is

$0.11 A$ .

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

Formula to calculate the capacitive reactance is,

$XC=12πfC$

$C$ is the capacitance of the capacitor
$f$ is the frequency of the ac source

Substitute $60 Hz$ for $f$ , $2.5 μF$ for $C$ to determine the capacitive reactance,

$XC=12π(60 Hz)(2.5 μF)=1.1×103 Ω$

The formula to calculate the impedance of the circuit where the components are in series is given by,

$Z=R2+(XL−XC)2$

$R$ is the resistance of the resistor
$XL$ is the inductive reactance
$XC$ is the capacitive reactance

Substitute $1.2 kΩ$ for $R$ , $0$ for $XL$ , $1.1×103 Ω$ for $XC$ to determine the impedance of the circuit,

$Z=(1.2 kΩ)2+(0−1.1×103 Ω)2=1.6×103 Ω$

Conclusion:

The maximum current in the circuit is given by,

$Im=VmZ$

$Vm$ is the maximum voltage supplied by the source
$Z$ is the impedance of the circuit

Substitute $1.6×103 Ω$ for $Z$ and $170 V$ for $Vm$ to determine the maximum current in the circuit,

$Im=170 V1.6×103 Ω=0.11 A$

The maximum current in the circuit is $0.11 A$ .

b)

Expert Solution

To determine

The maximum value of potential difference across the resistor and capacitor.

Answer to Problem 30P

The maximum voltage across the resistor is

$1.3×102 V$ and the capacitor is

$1.2×102 V$

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

Conclusion: The maximum voltage across the capacitor is $1.2×102 V$ .

c)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is

$0$ ,

$1.2×102 V$ ,

$1.2×102 V$ and

$300 μC$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum. When the current is minimum the voltage will be maximum.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $0$ for $ΔVR$ , $1.2×102 V$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=0+1.2×102 V=1.2×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is zero
$C$ is the capacitance

From above section the potential across the capacitor is when current is zero is $1.2×102 V$ .

Substitute $1.2×102 V$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(1.2×102 V)=(2.5×10−6 F)(1.2×102 V)=3.0×10−4 C=300 μC$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is $0$ , $1.2×102 V$ , $1.2×102 V$ and $300 μC$ respectively.

d)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is

$1.3×102 V$ ,

$0$ ,

$1.3×102 V$ , and

$0$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a resistor are in phase with each other. So when the current is maximum, the voltage across the resistor will be maximum.

The maximum potential across the resistor is

$Vm,R=ImR$

$Im$ is the maximum current in the circuit
$R$ is resistance

Substitute $0.11 A$ for $Im$ and $1.2 kΩ$ for $R$ to determine the maximum potential difference across the resistor,

$Vm,R=(0.11 A)(1.2 kΩ)=1.3×102 V$

The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the capacitor will reach its minimum.

Hence when the current reaches maximum, the voltage across the capacitor will be zero.

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $1.3×102 V$ for $ΔVR$ , $0$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=1.3×102 V+0=1.3×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is maximum
$C$ is the capacitance

From above section the potential across the capacitor is when current is maximum is $0$ .

Substitute $0$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(0)=0$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is $1.3×102 V$ , $0$ , $1.3×102 V$ , and $0$ respectively.

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Answer 2

Textbook Question

Chapter 21, Problem 30P

An AC source operating at 60. Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and a capacitor (C = 2.5 μF). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the capacitor? (c) When the current is zero, what are the magnitudes of the potential difference across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant? (d) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant?

(a)

Expert Solution

To determine

The maximum value of current in the circuit.

Answer to Problem 30P

The maximum current in the circuit is

$0.11 A$ .

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

Formula to calculate the capacitive reactance is,

$XC=12πfC$

$C$ is the capacitance of the capacitor
$f$ is the frequency of the ac source

Substitute $60 Hz$ for $f$ , $2.5 μF$ for $C$ to determine the capacitive reactance,

$XC=12π(60 Hz)(2.5 μF)=1.1×103 Ω$

The formula to calculate the impedance of the circuit where the components are in series is given by,

$Z=R2+(XL−XC)2$

$R$ is the resistance of the resistor
$XL$ is the inductive reactance
$XC$ is the capacitive reactance

Substitute $1.2 kΩ$ for $R$ , $0$ for $XL$ , $1.1×103 Ω$ for $XC$ to determine the impedance of the circuit,

$Z=(1.2 kΩ)2+(0−1.1×103 Ω)2=1.6×103 Ω$

Conclusion:

The maximum current in the circuit is given by,

$Im=VmZ$

$Vm$ is the maximum voltage supplied by the source
$Z$ is the impedance of the circuit

Substitute $1.6×103 Ω$ for $Z$ and $170 V$ for $Vm$ to determine the maximum current in the circuit,

$Im=170 V1.6×103 Ω=0.11 A$

The maximum current in the circuit is $0.11 A$ .

b)

Expert Solution

To determine

The maximum value of potential difference across the resistor and capacitor.

Answer to Problem 30P

The maximum voltage across the resistor is

$1.3×102 V$ and the capacitor is

$1.2×102 V$

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

Conclusion: The maximum voltage across the capacitor is $1.2×102 V$ .

c)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is

$0$ ,

$1.2×102 V$ ,

$1.2×102 V$ and

$300 μC$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum. When the current is minimum the voltage will be maximum.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $0$ for $ΔVR$ , $1.2×102 V$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=0+1.2×102 V=1.2×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is zero
$C$ is the capacitance

From above section the potential across the capacitor is when current is zero is $1.2×102 V$ .

Substitute $1.2×102 V$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(1.2×102 V)=(2.5×10−6 F)(1.2×102 V)=3.0×10−4 C=300 μC$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is $0$ , $1.2×102 V$ , $1.2×102 V$ and $300 μC$ respectively.

d)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is

$1.3×102 V$ ,

$0$ ,

$1.3×102 V$ , and

$0$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a resistor are in phase with each other. So when the current is maximum, the voltage across the resistor will be maximum.

The maximum potential across the resistor is

$Vm,R=ImR$

$Im$ is the maximum current in the circuit
$R$ is resistance

Substitute $0.11 A$ for $Im$ and $1.2 kΩ$ for $R$ to determine the maximum potential difference across the resistor,

$Vm,R=(0.11 A)(1.2 kΩ)=1.3×102 V$

The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the capacitor will reach its minimum.

Hence when the current reaches maximum, the voltage across the capacitor will be zero.

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $1.3×102 V$ for $ΔVR$ , $0$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=1.3×102 V+0=1.3×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is maximum
$C$ is the capacitance

From above section the potential across the capacitor is when current is maximum is $0$ .

Substitute $0$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(0)=0$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is $1.3×102 V$ , $0$ , $1.3×102 V$ , and $0$ respectively.

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Answer 3

Textbook Question

Chapter 21, Problem 30P

An AC source operating at 60. Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and a capacitor (C = 2.5 μF). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the capacitor? (c) When the current is zero, what are the magnitudes of the potential difference across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant? (d) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant?

(a)

Expert Solution

To determine

The maximum value of current in the circuit.

Answer to Problem 30P

The maximum current in the circuit is

$0.11 A$ .

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

Formula to calculate the capacitive reactance is,

$XC=12πfC$

$C$ is the capacitance of the capacitor
$f$ is the frequency of the ac source

Substitute $60 Hz$ for $f$ , $2.5 μF$ for $C$ to determine the capacitive reactance,

$XC=12π(60 Hz)(2.5 μF)=1.1×103 Ω$

The formula to calculate the impedance of the circuit where the components are in series is given by,

$Z=R2+(XL−XC)2$

$R$ is the resistance of the resistor
$XL$ is the inductive reactance
$XC$ is the capacitive reactance

Substitute $1.2 kΩ$ for $R$ , $0$ for $XL$ , $1.1×103 Ω$ for $XC$ to determine the impedance of the circuit,

$Z=(1.2 kΩ)2+(0−1.1×103 Ω)2=1.6×103 Ω$

Conclusion:

The maximum current in the circuit is given by,

$Im=VmZ$

$Vm$ is the maximum voltage supplied by the source
$Z$ is the impedance of the circuit

Substitute $1.6×103 Ω$ for $Z$ and $170 V$ for $Vm$ to determine the maximum current in the circuit,

$Im=170 V1.6×103 Ω=0.11 A$

The maximum current in the circuit is $0.11 A$ .

b)

Expert Solution

To determine

The maximum value of potential difference across the resistor and capacitor.

Answer to Problem 30P

The maximum voltage across the resistor is

$1.3×102 V$ and the capacitor is

$1.2×102 V$

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

Conclusion: The maximum voltage across the capacitor is $1.2×102 V$ .

c)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is

$0$ ,

$1.2×102 V$ ,

$1.2×102 V$ and

$300 μC$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum. When the current is minimum the voltage will be maximum.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $0$ for $ΔVR$ , $1.2×102 V$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=0+1.2×102 V=1.2×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is zero
$C$ is the capacitance

From above section the potential across the capacitor is when current is zero is $1.2×102 V$ .

Substitute $1.2×102 V$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(1.2×102 V)=(2.5×10−6 F)(1.2×102 V)=3.0×10−4 C=300 μC$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is $0$ , $1.2×102 V$ , $1.2×102 V$ and $300 μC$ respectively.

d)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is

$1.3×102 V$ ,

$0$ ,

$1.3×102 V$ , and

$0$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a resistor are in phase with each other. So when the current is maximum, the voltage across the resistor will be maximum.

The maximum potential across the resistor is

$Vm,R=ImR$

$Im$ is the maximum current in the circuit
$R$ is resistance

Substitute $0.11 A$ for $Im$ and $1.2 kΩ$ for $R$ to determine the maximum potential difference across the resistor,

$Vm,R=(0.11 A)(1.2 kΩ)=1.3×102 V$

The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the capacitor will reach its minimum.

Hence when the current reaches maximum, the voltage across the capacitor will be zero.

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $1.3×102 V$ for $ΔVR$ , $0$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=1.3×102 V+0=1.3×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is maximum
$C$ is the capacitance

From above section the potential across the capacitor is when current is maximum is $0$ .

Substitute $0$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(0)=0$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is $1.3×102 V$ , $0$ , $1.3×102 V$ , and $0$ respectively.

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Answer 4

Textbook Question

Chapter 21, Problem 30P

An AC source operating at 60. Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and a capacitor (C = 2.5 μF). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the capacitor? (c) When the current is zero, what are the magnitudes of the potential difference across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant? (d) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant?

(a)

Expert Solution

To determine

The maximum value of current in the circuit.

Answer to Problem 30P

The maximum current in the circuit is

$0.11 A$ .

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

Formula to calculate the capacitive reactance is,

$XC=12πfC$

$C$ is the capacitance of the capacitor
$f$ is the frequency of the ac source

Substitute $60 Hz$ for $f$ , $2.5 μF$ for $C$ to determine the capacitive reactance,

$XC=12π(60 Hz)(2.5 μF)=1.1×103 Ω$

The formula to calculate the impedance of the circuit where the components are in series is given by,

$Z=R2+(XL−XC)2$

$R$ is the resistance of the resistor
$XL$ is the inductive reactance
$XC$ is the capacitive reactance

Substitute $1.2 kΩ$ for $R$ , $0$ for $XL$ , $1.1×103 Ω$ for $XC$ to determine the impedance of the circuit,

$Z=(1.2 kΩ)2+(0−1.1×103 Ω)2=1.6×103 Ω$

Conclusion:

The maximum current in the circuit is given by,

$Im=VmZ$

$Vm$ is the maximum voltage supplied by the source
$Z$ is the impedance of the circuit

Substitute $1.6×103 Ω$ for $Z$ and $170 V$ for $Vm$ to determine the maximum current in the circuit,

$Im=170 V1.6×103 Ω=0.11 A$

The maximum current in the circuit is $0.11 A$ .

b)

Expert Solution

To determine

The maximum value of potential difference across the resistor and capacitor.

Answer to Problem 30P

The maximum voltage across the resistor is

$1.3×102 V$ and the capacitor is

$1.2×102 V$

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

Conclusion: The maximum voltage across the capacitor is $1.2×102 V$ .

c)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is

$0$ ,

$1.2×102 V$ ,

$1.2×102 V$ and

$300 μC$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum. When the current is minimum the voltage will be maximum.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $0$ for $ΔVR$ , $1.2×102 V$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=0+1.2×102 V=1.2×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is zero
$C$ is the capacitance

From above section the potential across the capacitor is when current is zero is $1.2×102 V$ .

Substitute $1.2×102 V$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(1.2×102 V)=(2.5×10−6 F)(1.2×102 V)=3.0×10−4 C=300 μC$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is $0$ , $1.2×102 V$ , $1.2×102 V$ and $300 μC$ respectively.

d)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is

$1.3×102 V$ ,

$0$ ,

$1.3×102 V$ , and

$0$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a resistor are in phase with each other. So when the current is maximum, the voltage across the resistor will be maximum.

The maximum potential across the resistor is

$Vm,R=ImR$

$Im$ is the maximum current in the circuit
$R$ is resistance

Substitute $0.11 A$ for $Im$ and $1.2 kΩ$ for $R$ to determine the maximum potential difference across the resistor,

$Vm,R=(0.11 A)(1.2 kΩ)=1.3×102 V$

The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the capacitor will reach its minimum.

Hence when the current reaches maximum, the voltage across the capacitor will be zero.

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $1.3×102 V$ for $ΔVR$ , $0$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=1.3×102 V+0=1.3×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is maximum
$C$ is the capacitance

From above section the potential across the capacitor is when current is maximum is $0$ .

Substitute $0$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(0)=0$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is $1.3×102 V$ , $0$ , $1.3×102 V$ , and $0$ respectively.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The maximum value of current in the circuit.

Answer to Problem 30P

The maximum current in the circuit is

$0.11 A$ .

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

Formula to calculate the capacitive reactance is,

$XC=12πfC$

$C$ is the capacitance of the capacitor
$f$ is the frequency of the ac source

Substitute $60 Hz$ for $f$ , $2.5 μF$ for $C$ to determine the capacitive reactance,

$XC=12π(60 Hz)(2.5 μF)=1.1×103 Ω$

The formula to calculate the impedance of the circuit where the components are in series is given by,

$Z=R2+(XL−XC)2$

$R$ is the resistance of the resistor
$XL$ is the inductive reactance
$XC$ is the capacitive reactance

Substitute $1.2 kΩ$ for $R$ , $0$ for $XL$ , $1.1×103 Ω$ for $XC$ to determine the impedance of the circuit,

$Z=(1.2 kΩ)2+(0−1.1×103 Ω)2=1.6×103 Ω$

Conclusion:

The maximum current in the circuit is given by,

$Im=VmZ$

$Vm$ is the maximum voltage supplied by the source
$Z$ is the impedance of the circuit

Substitute $1.6×103 Ω$ for $Z$ and $170 V$ for $Vm$ to determine the maximum current in the circuit,

$Im=170 V1.6×103 Ω=0.11 A$

The maximum current in the circuit is $0.11 A$ .

b)

Expert Solution

To determine

The maximum value of potential difference across the resistor and capacitor.

Answer to Problem 30P

The maximum voltage across the resistor is

$1.3×102 V$ and the capacitor is

$1.2×102 V$

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

Conclusion: The maximum voltage across the capacitor is $1.2×102 V$ .

c)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is

$0$ ,

$1.2×102 V$ ,

$1.2×102 V$ and

$300 μC$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum. When the current is minimum the voltage will be maximum.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $0$ for $ΔVR$ , $1.2×102 V$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=0+1.2×102 V=1.2×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is zero
$C$ is the capacitance

From above section the potential across the capacitor is when current is zero is $1.2×102 V$ .

Substitute $1.2×102 V$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(1.2×102 V)=(2.5×10−6 F)(1.2×102 V)=3.0×10−4 C=300 μC$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is $0$ , $1.2×102 V$ , $1.2×102 V$ and $300 μC$ respectively.

d)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is

$1.3×102 V$ ,

$0$ ,

$1.3×102 V$ , and

$0$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a resistor are in phase with each other. So when the current is maximum, the voltage across the resistor will be maximum.

The maximum potential across the resistor is

$Vm,R=ImR$

$Im$ is the maximum current in the circuit
$R$ is resistance

Substitute $0.11 A$ for $Im$ and $1.2 kΩ$ for $R$ to determine the maximum potential difference across the resistor,

$Vm,R=(0.11 A)(1.2 kΩ)=1.3×102 V$

The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the capacitor will reach its minimum.

Hence when the current reaches maximum, the voltage across the capacitor will be zero.

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $1.3×102 V$ for $ΔVR$ , $0$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=1.3×102 V+0=1.3×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is maximum
$C$ is the capacitance

From above section the potential across the capacitor is when current is maximum is $0$ .

Substitute $0$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(0)=0$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is $1.3×102 V$ , $0$ , $1.3×102 V$ , and $0$ respectively.

Answer 8

(a)

Expert Solution

To determine

The maximum value of current in the circuit.

Answer to Problem 30P

The maximum current in the circuit is

$0.11 A$ .

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

Formula to calculate the capacitive reactance is,

$XC=12πfC$

$C$ is the capacitance of the capacitor
$f$ is the frequency of the ac source

Substitute $60 Hz$ for $f$ , $2.5 μF$ for $C$ to determine the capacitive reactance,

$XC=12π(60 Hz)(2.5 μF)=1.1×103 Ω$

The formula to calculate the impedance of the circuit where the components are in series is given by,

$Z=R2+(XL−XC)2$

$R$ is the resistance of the resistor
$XL$ is the inductive reactance
$XC$ is the capacitive reactance

Substitute $1.2 kΩ$ for $R$ , $0$ for $XL$ , $1.1×103 Ω$ for $XC$ to determine the impedance of the circuit,

$Z=(1.2 kΩ)2+(0−1.1×103 Ω)2=1.6×103 Ω$

Conclusion:

The maximum current in the circuit is given by,

$Im=VmZ$

$Vm$ is the maximum voltage supplied by the source
$Z$ is the impedance of the circuit

Substitute $1.6×103 Ω$ for $Z$ and $170 V$ for $Vm$ to determine the maximum current in the circuit,

$Im=170 V1.6×103 Ω=0.11 A$

The maximum current in the circuit is $0.11 A$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The maximum value of current in the circuit.

Answer 13

Answer to Problem 30P

The maximum current in the circuit is

$0.11 A$ .

Answer 14

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

Formula to calculate the capacitive reactance is,

$XC=12πfC$

$C$ is the capacitance of the capacitor
$f$ is the frequency of the ac source

Substitute $60 Hz$ for $f$ , $2.5 μF$ for $C$ to determine the capacitive reactance,

$XC=12π(60 Hz)(2.5 μF)=1.1×103 Ω$

The formula to calculate the impedance of the circuit where the components are in series is given by,

$Z=R2+(XL−XC)2$

$R$ is the resistance of the resistor
$XL$ is the inductive reactance
$XC$ is the capacitive reactance

Substitute $1.2 kΩ$ for $R$ , $0$ for $XL$ , $1.1×103 Ω$ for $XC$ to determine the impedance of the circuit,

$Z=(1.2 kΩ)2+(0−1.1×103 Ω)2=1.6×103 Ω$

Conclusion:

The maximum current in the circuit is given by,

$Im=VmZ$

$Vm$ is the maximum voltage supplied by the source
$Z$ is the impedance of the circuit

Substitute $1.6×103 Ω$ for $Z$ and $170 V$ for $Vm$ to determine the maximum current in the circuit,

$Im=170 V1.6×103 Ω=0.11 A$

The maximum current in the circuit is $0.11 A$ .

Answer 15

b)

Expert Solution

To determine

The maximum value of potential difference across the resistor and capacitor.

Answer to Problem 30P

The maximum voltage across the resistor is

$1.3×102 V$ and the capacitor is

$1.2×102 V$

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

Conclusion: The maximum voltage across the capacitor is $1.2×102 V$ .

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The maximum value of potential difference across the resistor and capacitor.

Answer 20

Answer to Problem 30P

The maximum voltage across the resistor is

$1.3×102 V$ and the capacitor is

$1.2×102 V$

Answer 21

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

Conclusion: The maximum voltage across the capacitor is $1.2×102 V$ .

Answer 22

c)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is

$0$ ,

$1.2×102 V$ ,

$1.2×102 V$ and

$300 μC$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum. When the current is minimum the voltage will be maximum.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $0$ for $ΔVR$ , $1.2×102 V$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=0+1.2×102 V=1.2×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is zero
$C$ is the capacitance

From above section the potential across the capacitor is when current is zero is $1.2×102 V$ .

Substitute $1.2×102 V$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(1.2×102 V)=(2.5×10−6 F)(1.2×102 V)=3.0×10−4 C=300 μC$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is $0$ , $1.2×102 V$ , $1.2×102 V$ and $300 μC$ respectively.

Answer 23

c)

Expert Solution

Answer 24

c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant.

Answer 27

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is

$0$ ,

$1.2×102 V$ ,

$1.2×102 V$ and

$300 μC$ respectively.

Answer 28

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the inductor will reach its minimum. When the current is minimum the voltage will be maximum.

The maximum potential across the capacitor is

$Vm,C=ImXC$

$Im$ is the maximum current in the circuit
$XC$ is capacitive reactance

Substitute $0.11 A$ for $Im$ and $1.1×103 Ω$ for $XC$ to determine the maximum potential difference across the resistor,

$Vm,C=(0.11 A)(1.1×103 Ω)=1.2×102 V$

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $0$ for $ΔVR$ , $1.2×102 V$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=0+1.2×102 V=1.2×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is zero
$C$ is the capacitance

From above section the potential across the capacitor is when current is zero is $1.2×102 V$ .

Substitute $1.2×102 V$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(1.2×102 V)=(2.5×10−6 F)(1.2×102 V)=3.0×10−4 C=300 μC$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is zero and the charge on the capacitor at that instant is $0$ , $1.2×102 V$ , $1.2×102 V$ and $300 μC$ respectively.

Answer 29

d)

Expert Solution

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant.

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is

$1.3×102 V$ ,

$0$ ,

$1.3×102 V$ , and

$0$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a resistor are in phase with each other. So when the current is maximum, the voltage across the resistor will be maximum.

The maximum potential across the resistor is

$Vm,R=ImR$

$Im$ is the maximum current in the circuit
$R$ is resistance

Substitute $0.11 A$ for $Im$ and $1.2 kΩ$ for $R$ to determine the maximum potential difference across the resistor,

$Vm,R=(0.11 A)(1.2 kΩ)=1.3×102 V$

The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the capacitor will reach its minimum.

Hence when the current reaches maximum, the voltage across the capacitor will be zero.

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $1.3×102 V$ for $ΔVR$ , $0$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=1.3×102 V+0=1.3×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is maximum
$C$ is the capacitance

From above section the potential across the capacitor is when current is maximum is $0$ .

Substitute $0$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(0)=0$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is $1.3×102 V$ , $0$ , $1.3×102 V$ , and $0$ respectively.

Answer 30

d)

Expert Solution

Answer 31

d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant.

Answer 34

Answer to Problem 30P

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is

$1.3×102 V$ ,

$0$ ,

$1.3×102 V$ , and

$0$ respectively.

Answer 35

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a resistor are in phase with each other. So when the current is maximum, the voltage across the resistor will be maximum.

The maximum potential across the resistor is

$Vm,R=ImR$

$Im$ is the maximum current in the circuit
$R$ is resistance

Substitute $0.11 A$ for $Im$ and $1.2 kΩ$ for $R$ to determine the maximum potential difference across the resistor,

$Vm,R=(0.11 A)(1.2 kΩ)=1.3×102 V$

The maximum voltage of the AC source is $170 V$ and has a frequency of $60 Hz$ . A resistor of resistance $1.2 kΩ$ and a capacitor of capacitance $2.5 μF$ are connected in series in the circuit.

The current and the voltage across a capacitor are $90°$ out of phase with each other. Hence when the current reaches its maximum, the voltage across the capacitor will reach its minimum.

Hence when the current reaches maximum, the voltage across the capacitor will be zero.

When Kirchoff’s law is applied to the closed circuit,

$ΔVs−ΔVR−ΔVC=0$

$ΔVs$ is the voltage across the AC source
$ΔVR$ is the voltage across the resistor
$ΔVC$ is the voltage across the capacitor

The voltage across the AC source will be,

$ΔVs=ΔVR+ΔVC$

Substitute $1.3×102 V$ for $ΔVR$ , $0$ for $ΔVC$ to determine the voltage across the AC source,

$ΔVs=1.3×102 V+0=1.3×102 V$

The charge accumulated in the capacitor is given by,

$q=CVC$

$VC$ is the potential across the capacitor at the instant the current is maximum
$C$ is the capacitance

From above section the potential across the capacitor is when current is maximum is $0$ .

Substitute $0$ for $VC$ and $2.5 μF$ for $C$ to determine the charge across the capacitor,

$q=(2.5 μF)(0)=0$

Conclusion:

The magnitudes of potential difference across the resistor, the capacitor, and the AC source when the current is maximum and the charge on the capacitor at that instant is $1.3×102 V$ , $0$ , $1.3×102 V$ , and $0$ respectively.