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?

To determine

The maximum value of current in the circuit.

Answer to Problem 30P

The maximum current in the circuit is $0.11\text{A}$.

Explanation of Solution

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

Formula to calculate the capacitive reactance is,

$X_C=\frac{1}{2\pi fC}$

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

Substitute $60\text{Hz}$ for $f$, $2.5\text{μF}$ for $C$ to determine the capacitive reactance,

$\begin{array}{c}{X}_C=\frac{1}{2\pi \left(60\text{Hz}\right)\left(2.5\text{μF}\right)}\\ =1.1\times {10}^3\text{Ω}\end{array}$

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

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

• $R$ is the resistance of the resistor
• $X_L$ is the inductive reactance
• $X_C$ is the capacitive reactance

Substitute $1.2\text{kΩ}$ for $R$, $0$ for $X_L$, $1.1\times {10}^3\text{Ω}$ for $X_C$ to determine the impedance of the circuit,

$\begin{array}{c}Z=\sqrt{{\left(1.2\text{kΩ}\right)}^2+{\left(0-1.1\times {10}^3\text{Ω}\right)}^2}\\ =1.6\times {10}^3\text{Ω}\end{array}$

Conclusion:

The maximum current in the circuit is given by,

$I_m=\frac{V_m}{Z}$

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

Substitute $1.6\times {10}^3\text{Ω}$ for $Z$ and $170\text{V}$ for $V_m$ to determine the maximum current in the circuit,

$\begin{array}{c}{I}_m=\frac{170\text{V}}{1.6\times {10}^3\text{Ω}}\\ =0.11\text{A}\end{array}$

The maximum current in the circuit is $0.11\text{A}$.

To determine

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

Answer to Problem 30P

The maximum voltage across the resistor is $1.3\times {10}^2\text{V}$ and the capacitor is $1.2\times {10}^2\text{V}$

Explanation of Solution

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

The maximum potential across the capacitor is

$V_{m,C}=I_m{X}_C$

• $I_m$ is the maximum current in the circuit
• $X_C$ is capacitive reactance

Substitute $0.11\text{A}$ for $I_m$ and $1.1\times {10}^3\text{Ω}$ for $X_C$ to determine the maximum potential difference across the resistor,

$\begin{array}{c}{V}_{m,C}=\left(0.11\text{A}\right)\left(1.1\times {10}^3\text{Ω}\right)\\ =1.2\times {10}^2\text{V}\end{array}$

Conclusion: The maximum voltage across the capacitor is $1.2\times {10}^2\text{V}$.

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 $0$, $1.2\times {10}^2\text{V}$, $1.2\times {10}^2\text{V}$ and $\text{300}\text{μC}$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170\text{V}$ and has a frequency of $60\text{Hz}$. A resistor of resistance $1.2\text{kΩ}$ and a capacitor of capacitance $2.5\text{μF}$ 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

$V_{m,C}=I_m{X}_C$

• $I_m$ is the maximum current in the circuit
• $X_C$ is capacitive reactance

Substitute $0.11\text{A}$ for $I_m$ and $1.1\times {10}^3\text{Ω}$ for $X_C$ to determine the maximum potential difference across the resistor,

$\begin{array}{c}{V}_{m,C}=\left(0.11\text{A}\right)\left(1.1\times {10}^3\text{Ω}\right)\\ =1.2\times {10}^2\text{V}\end{array}$

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

$\Delta V_s-\Delta V_R-\Delta V_C=0$

• $\Delta V_s$ is the voltage across the AC source
• $\Delta V_R$ is the voltage across the resistor
• $\Delta V_C$ is the voltage across the capacitor

The voltage across the AC source will be,

$\Delta V_s=\Delta V_R+\Delta V_C$

Substitute $0$ for $\Delta V_R$, $1.2\times {10}^2\text{V}$ for $\Delta V_C$ to determine the voltage across the AC source,

$\begin{array}{c}\Delta V_s=0+1.2\times {10}^2\text{V}\\ =1.2\times {10}^2\text{V}\end{array}$

The charge accumulated in the capacitor is given by,

$q=C{V}_C$

• $V_C$ 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 $1.2\times {10}^2\text{V}$.

Substitute $1.2\times {10}^2\text{V}$ for $V_C$ and $2.5\text{μF}$ for $C$ to determine the charge across the capacitor,

$\begin{array}{c}q=\left(2.5\text{μF}\right)\left(1.2\times {10}^2\text{V}\right)\\ =\left(2.5\times {10}^{-6}\text{F}\right)\left(1.2\times {10}^2\text{V}\right)\\ =3.0\times {10}^{-4}\text{C}\\ \text{=300}\text{μC}\end{array}$

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 $0$, $1.2\times {10}^2\text{V}$, $1.2\times {10}^2\text{V}$ and $\text{300}\text{μC}$ respectively.

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 $1.3\times {10}^2\text{V}$, $0$ , $1.3\times {10}^2\text{V}$, and $0$ respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is $170\text{V}$ and has a frequency of $60\text{Hz}$. A resistor of resistance $1.2\text{kΩ}$ and a capacitor of capacitance $2.5\text{μF}$ 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

$V_{m,R}=I_{m}R$

• $I_m$ is the maximum current in the circuit
• $R$ is resistance

Substitute $0.11\text{A}$ for $I_m$ and $1.2\text{kΩ}$ for $R$ to determine the maximum potential difference across the resistor,

$\begin{array}{c}{V}_{m,R}=\left(0.11\text{A}\right)\left(1.2\text{kΩ}\right)\\ =1.3\times {10}^2\text{V}\end{array}$

The maximum voltage of the AC source is $170\text{V}$ and has a frequency of $60\text{Hz}$. A resistor of resistance $1.2\text{kΩ}$ and a capacitor of capacitance $2.5\text{μF}$ 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,

$\Delta V_s-\Delta V_R-\Delta V_C=0$

• $\Delta V_s$ is the voltage across the AC source
• $\Delta V_R$ is the voltage across the resistor
• $\Delta V_C$ is the voltage across the capacitor

The voltage across the AC source will be,

$\Delta V_s=\Delta V_R+\Delta V_C$

Substitute $1.3\times {10}^2\text{V}$ for $\Delta V_R$, $0$ for $\Delta V_C$ to determine the voltage across the AC source,

$\begin{array}{c}\Delta V_s=1.3\times {10}^2\text{V}+0\\ =1.3\times {10}^2\text{V}\end{array}$

The charge accumulated in the capacitor is given by,

$q=C{V}_C$

• $V_C$ 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 $V_C$ and $2.5\text{μF}$ for $C$ to determine the charge across the capacitor,

$\begin{array}{c}q=\left(2.5\text{μF}\right)\left(0\right)\\ =0\end{array}$

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 $1.3\times {10}^2\text{V}$, $0$ , $1.3\times {10}^2\text{V}$, and $0$ respectively.