The average power supplied to the circuit.

Question

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Answer 1

Question

Chapter 29, Problem 70P

(a)

To determine

The average power supplied to the circuit.

(a)

Expert Solution

Answer to Problem 70P

$Pav=933.38W$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Calculation:

The average power is

$Pav=VrmsIrmscosϕPav=120V×11A×cos 45oPav=933.38W$

Conclusion:

The average power supplied to the circuit is $Pav=933.38W$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 70P

$R=7.71Ω$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Calculation:

The resistance is,

$Pav=Irms2RR=P avI rms2R=933.38W ( 11A )2R=7.71Ω$

Conclusion:

$R=7.71Ω$

This is the resistance of the circuit.

(c)

To determine

The capacitance of the circuit.

(c)

Expert Solution

Answer to Problem 70P

$C=2.4×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Resistance, $R=7.71Ω$

Formula used:

Capacitive reactance, $XC=12πfC$

Impedance, $Z2=R2+(XL−XC)2=ε2Irms2$

Calculation:

The capacitive impedance is,

$Z2=R2+( X L− X C)2ε2I rms2=R2+(2πfL− X C)2( 120V 11A)2=7.71Ω2+(2π×60Hz×50× 10 -3− X C)2119.01=59.44+(18.85− X C)218.85−XC=119.01−59.44XC=18.85−7.71XC=11.14 Ω$

The capacitance is derived as,

$C=12πfXCC=12π×60Hz×11.14 ΩC=2.4×10−4F$

Conclusion:

The capacitance is $C=2.4×10−4F$ .

(d)

To determine

The change in the capacitance to make the power factor equal to 1.

(d)

Expert Solution

Answer to Problem 70P

$ΔC=1×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Cpf=1$ .The frequency is independent on the power factor and the inductance is remaining constant as per the conditions. So, $Cpf=1$ is establish as,

$CP f=1=1 ( 2πf )2LCP f=1=1 ( 2π×60Hz )2×0.05HCP f=1=1.4×10−4F$

The change in the capacitance for which the power factor is 1 is,

$ΔC=C−CP f=1ΔC=(2.4× 10 −4−1.4× 10 −4)FΔC=1×10−4F$

Conclusion:

The capacitance must change $ΔC=1×10−4F$

(e)

To determine

The change in inductance to make the power factor 1.

(e)

Expert Solution

Answer to Problem 70P

$ΔL=0.02H$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH=0.05H$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Lpf=1$ . The frequency is independent of the power factor and the capacitance is remains constant as per the conditions. So, $Lpf=1$ is established as:

$LP f=1=1 ( 2πf )2CLP f=1=1 ( 2π×60Hz )2×2.4× 10 −4FLP f=1=0.03 H$

The change in the inductance for which the power factor is 1 is

$ΔL=L−LP f=1ΔL=(0.05−0.03) HΔL=0.02H$

Conclusion:

The change in inductance to make power factor 1 is $ΔL=0.02H$ .

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Answer 2

Question

Chapter 29, Problem 70P

(a)

To determine

The average power supplied to the circuit.

(a)

Expert Solution

Answer to Problem 70P

$Pav=933.38W$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Calculation:

The average power is

$Pav=VrmsIrmscosϕPav=120V×11A×cos 45oPav=933.38W$

Conclusion:

The average power supplied to the circuit is $Pav=933.38W$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 70P

$R=7.71Ω$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Calculation:

The resistance is,

$Pav=Irms2RR=P avI rms2R=933.38W ( 11A )2R=7.71Ω$

Conclusion:

$R=7.71Ω$

This is the resistance of the circuit.

(c)

To determine

The capacitance of the circuit.

(c)

Expert Solution

Answer to Problem 70P

$C=2.4×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Resistance, $R=7.71Ω$

Formula used:

Capacitive reactance, $XC=12πfC$

Impedance, $Z2=R2+(XL−XC)2=ε2Irms2$

Calculation:

The capacitive impedance is,

$Z2=R2+( X L− X C)2ε2I rms2=R2+(2πfL− X C)2( 120V 11A)2=7.71Ω2+(2π×60Hz×50× 10 -3− X C)2119.01=59.44+(18.85− X C)218.85−XC=119.01−59.44XC=18.85−7.71XC=11.14 Ω$

The capacitance is derived as,

$C=12πfXCC=12π×60Hz×11.14 ΩC=2.4×10−4F$

Conclusion:

The capacitance is $C=2.4×10−4F$ .

(d)

To determine

The change in the capacitance to make the power factor equal to 1.

(d)

Expert Solution

Answer to Problem 70P

$ΔC=1×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Cpf=1$ .The frequency is independent on the power factor and the inductance is remaining constant as per the conditions. So, $Cpf=1$ is establish as,

$CP f=1=1 ( 2πf )2LCP f=1=1 ( 2π×60Hz )2×0.05HCP f=1=1.4×10−4F$

The change in the capacitance for which the power factor is 1 is,

$ΔC=C−CP f=1ΔC=(2.4× 10 −4−1.4× 10 −4)FΔC=1×10−4F$

Conclusion:

The capacitance must change $ΔC=1×10−4F$

(e)

To determine

The change in inductance to make the power factor 1.

(e)

Expert Solution

Answer to Problem 70P

$ΔL=0.02H$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH=0.05H$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Lpf=1$ . The frequency is independent of the power factor and the capacitance is remains constant as per the conditions. So, $Lpf=1$ is established as:

$LP f=1=1 ( 2πf )2CLP f=1=1 ( 2π×60Hz )2×2.4× 10 −4FLP f=1=0.03 H$

The change in the inductance for which the power factor is 1 is

$ΔL=L−LP f=1ΔL=(0.05−0.03) HΔL=0.02H$

Conclusion:

The change in inductance to make power factor 1 is $ΔL=0.02H$ .

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Answer 3

Question

Chapter 29, Problem 70P

(a)

To determine

The average power supplied to the circuit.

(a)

Expert Solution

Answer to Problem 70P

$Pav=933.38W$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Calculation:

The average power is

$Pav=VrmsIrmscosϕPav=120V×11A×cos 45oPav=933.38W$

Conclusion:

The average power supplied to the circuit is $Pav=933.38W$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 70P

$R=7.71Ω$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Calculation:

The resistance is,

$Pav=Irms2RR=P avI rms2R=933.38W ( 11A )2R=7.71Ω$

Conclusion:

$R=7.71Ω$

This is the resistance of the circuit.

(c)

To determine

The capacitance of the circuit.

(c)

Expert Solution

Answer to Problem 70P

$C=2.4×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Resistance, $R=7.71Ω$

Formula used:

Capacitive reactance, $XC=12πfC$

Impedance, $Z2=R2+(XL−XC)2=ε2Irms2$

Calculation:

The capacitive impedance is,

$Z2=R2+( X L− X C)2ε2I rms2=R2+(2πfL− X C)2( 120V 11A)2=7.71Ω2+(2π×60Hz×50× 10 -3− X C)2119.01=59.44+(18.85− X C)218.85−XC=119.01−59.44XC=18.85−7.71XC=11.14 Ω$

The capacitance is derived as,

$C=12πfXCC=12π×60Hz×11.14 ΩC=2.4×10−4F$

Conclusion:

The capacitance is $C=2.4×10−4F$ .

(d)

To determine

The change in the capacitance to make the power factor equal to 1.

(d)

Expert Solution

Answer to Problem 70P

$ΔC=1×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Cpf=1$ .The frequency is independent on the power factor and the inductance is remaining constant as per the conditions. So, $Cpf=1$ is establish as,

$CP f=1=1 ( 2πf )2LCP f=1=1 ( 2π×60Hz )2×0.05HCP f=1=1.4×10−4F$

The change in the capacitance for which the power factor is 1 is,

$ΔC=C−CP f=1ΔC=(2.4× 10 −4−1.4× 10 −4)FΔC=1×10−4F$

Conclusion:

The capacitance must change $ΔC=1×10−4F$

(e)

To determine

The change in inductance to make the power factor 1.

(e)

Expert Solution

Answer to Problem 70P

$ΔL=0.02H$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH=0.05H$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Lpf=1$ . The frequency is independent of the power factor and the capacitance is remains constant as per the conditions. So, $Lpf=1$ is established as:

$LP f=1=1 ( 2πf )2CLP f=1=1 ( 2π×60Hz )2×2.4× 10 −4FLP f=1=0.03 H$

The change in the inductance for which the power factor is 1 is

$ΔL=L−LP f=1ΔL=(0.05−0.03) HΔL=0.02H$

Conclusion:

The change in inductance to make power factor 1 is $ΔL=0.02H$ .

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Answer 4

Question

Chapter 29, Problem 70P

(a)

To determine

The average power supplied to the circuit.

(a)

Expert Solution

Answer to Problem 70P

$Pav=933.38W$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Calculation:

The average power is

$Pav=VrmsIrmscosϕPav=120V×11A×cos 45oPav=933.38W$

Conclusion:

The average power supplied to the circuit is $Pav=933.38W$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 70P

$R=7.71Ω$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Calculation:

The resistance is,

$Pav=Irms2RR=P avI rms2R=933.38W ( 11A )2R=7.71Ω$

Conclusion:

$R=7.71Ω$

This is the resistance of the circuit.

(c)

To determine

The capacitance of the circuit.

(c)

Expert Solution

Answer to Problem 70P

$C=2.4×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Resistance, $R=7.71Ω$

Formula used:

Capacitive reactance, $XC=12πfC$

Impedance, $Z2=R2+(XL−XC)2=ε2Irms2$

Calculation:

The capacitive impedance is,

$Z2=R2+( X L− X C)2ε2I rms2=R2+(2πfL− X C)2( 120V 11A)2=7.71Ω2+(2π×60Hz×50× 10 -3− X C)2119.01=59.44+(18.85− X C)218.85−XC=119.01−59.44XC=18.85−7.71XC=11.14 Ω$

The capacitance is derived as,

$C=12πfXCC=12π×60Hz×11.14 ΩC=2.4×10−4F$

Conclusion:

The capacitance is $C=2.4×10−4F$ .

(d)

To determine

The change in the capacitance to make the power factor equal to 1.

(d)

Expert Solution

Answer to Problem 70P

$ΔC=1×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Cpf=1$ .The frequency is independent on the power factor and the inductance is remaining constant as per the conditions. So, $Cpf=1$ is establish as,

$CP f=1=1 ( 2πf )2LCP f=1=1 ( 2π×60Hz )2×0.05HCP f=1=1.4×10−4F$

The change in the capacitance for which the power factor is 1 is,

$ΔC=C−CP f=1ΔC=(2.4× 10 −4−1.4× 10 −4)FΔC=1×10−4F$

Conclusion:

The capacitance must change $ΔC=1×10−4F$

(e)

To determine

The change in inductance to make the power factor 1.

(e)

Expert Solution

Answer to Problem 70P

$ΔL=0.02H$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH=0.05H$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Lpf=1$ . The frequency is independent of the power factor and the capacitance is remains constant as per the conditions. So, $Lpf=1$ is established as:

$LP f=1=1 ( 2πf )2CLP f=1=1 ( 2π×60Hz )2×2.4× 10 −4FLP f=1=0.03 H$

The change in the inductance for which the power factor is 1 is

$ΔL=L−LP f=1ΔL=(0.05−0.03) HΔL=0.02H$

Conclusion:

The change in inductance to make power factor 1 is $ΔL=0.02H$ .

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Answer 5

Chapter 29, Problem 70P

(a)

To determine

The average power supplied to the circuit.

(a)

Expert Solution

Answer to Problem 70P

$Pav=933.38W$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Calculation:

The average power is

$Pav=VrmsIrmscosϕPav=120V×11A×cos 45oPav=933.38W$

Conclusion:

The average power supplied to the circuit is $Pav=933.38W$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 70P

$R=7.71Ω$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Calculation:

The resistance is,

$Pav=Irms2RR=P avI rms2R=933.38W ( 11A )2R=7.71Ω$

Conclusion:

$R=7.71Ω$

This is the resistance of the circuit.

(c)

To determine

The capacitance of the circuit.

(c)

Expert Solution

Answer to Problem 70P

$C=2.4×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Resistance, $R=7.71Ω$

Formula used:

Capacitive reactance, $XC=12πfC$

Impedance, $Z2=R2+(XL−XC)2=ε2Irms2$

Calculation:

The capacitive impedance is,

$Z2=R2+( X L− X C)2ε2I rms2=R2+(2πfL− X C)2( 120V 11A)2=7.71Ω2+(2π×60Hz×50× 10 -3− X C)2119.01=59.44+(18.85− X C)218.85−XC=119.01−59.44XC=18.85−7.71XC=11.14 Ω$

The capacitance is derived as,

$C=12πfXCC=12π×60Hz×11.14 ΩC=2.4×10−4F$

Conclusion:

The capacitance is $C=2.4×10−4F$ .

(d)

To determine

The change in the capacitance to make the power factor equal to 1.

(d)

Expert Solution

Answer to Problem 70P

$ΔC=1×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Cpf=1$ .The frequency is independent on the power factor and the inductance is remaining constant as per the conditions. So, $Cpf=1$ is establish as,

$CP f=1=1 ( 2πf )2LCP f=1=1 ( 2π×60Hz )2×0.05HCP f=1=1.4×10−4F$

The change in the capacitance for which the power factor is 1 is,

$ΔC=C−CP f=1ΔC=(2.4× 10 −4−1.4× 10 −4)FΔC=1×10−4F$

Conclusion:

The capacitance must change $ΔC=1×10−4F$

(e)

To determine

The change in inductance to make the power factor 1.

(e)

Expert Solution

Answer to Problem 70P

$ΔL=0.02H$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH=0.05H$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Lpf=1$ . The frequency is independent of the power factor and the capacitance is remains constant as per the conditions. So, $Lpf=1$ is established as:

$LP f=1=1 ( 2πf )2CLP f=1=1 ( 2π×60Hz )2×2.4× 10 −4FLP f=1=0.03 H$

The change in the inductance for which the power factor is 1 is

$ΔL=L−LP f=1ΔL=(0.05−0.03) HΔL=0.02H$

Conclusion:

The change in inductance to make power factor 1 is $ΔL=0.02H$ .

Answer 6

Chapter 29, Problem 70P

(a)

To determine

The average power supplied to the circuit.

(a)

Expert Solution

Answer to Problem 70P

$Pav=933.38W$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Calculation:

The average power is

$Pav=VrmsIrmscosϕPav=120V×11A×cos 45oPav=933.38W$

Conclusion:

The average power supplied to the circuit is $Pav=933.38W$ .

Answer 7

Chapter 29, Problem 70P

(a)

To determine

The average power supplied to the circuit.

Answer 8

(a)

Expert Solution

Answer to Problem 70P

$Pav=933.38W$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Calculation:

The average power is

$Pav=VrmsIrmscosϕPav=120V×11A×cos 45oPav=933.38W$

Conclusion:

The average power supplied to the circuit is $Pav=933.38W$ .

Answer 13

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 70P

$R=7.71Ω$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Calculation:

The resistance is,

$Pav=Irms2RR=P avI rms2R=933.38W ( 11A )2R=7.71Ω$

Conclusion:

$R=7.71Ω$

This is the resistance of the circuit.

Answer 14

(b)

To determine

The resistance of the circuit.

Answer 15

(b)

Expert Solution

Answer to Problem 70P

$R=7.71Ω$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Calculation:

The resistance is,

$Pav=Irms2RR=P avI rms2R=933.38W ( 11A )2R=7.71Ω$

Conclusion:

$R=7.71Ω$

This is the resistance of the circuit.

Answer 20

(c)

To determine

The capacitance of the circuit.

(c)

Expert Solution

Answer to Problem 70P

$C=2.4×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Resistance, $R=7.71Ω$

Formula used:

Capacitive reactance, $XC=12πfC$

Impedance, $Z2=R2+(XL−XC)2=ε2Irms2$

Calculation:

The capacitive impedance is,

$Z2=R2+( X L− X C)2ε2I rms2=R2+(2πfL− X C)2( 120V 11A)2=7.71Ω2+(2π×60Hz×50× 10 -3− X C)2119.01=59.44+(18.85− X C)218.85−XC=119.01−59.44XC=18.85−7.71XC=11.14 Ω$

The capacitance is derived as,

$C=12πfXCC=12π×60Hz×11.14 ΩC=2.4×10−4F$

Conclusion:

The capacitance is $C=2.4×10−4F$ .

Answer 21

(c)

To determine

The capacitance of the circuit.

Answer 22

(c)

Expert Solution

Answer to Problem 70P

$C=2.4×10−4F$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Resistance, $R=7.71Ω$

Formula used:

Capacitive reactance, $XC=12πfC$

Impedance, $Z2=R2+(XL−XC)2=ε2Irms2$

Calculation:

The capacitive impedance is,

$Z2=R2+( X L− X C)2ε2I rms2=R2+(2πfL− X C)2( 120V 11A)2=7.71Ω2+(2π×60Hz×50× 10 -3− X C)2119.01=59.44+(18.85− X C)218.85−XC=119.01−59.44XC=18.85−7.71XC=11.14 Ω$

The capacitance is derived as,

$C=12πfXCC=12π×60Hz×11.14 ΩC=2.4×10−4F$

Conclusion:

The capacitance is $C=2.4×10−4F$ .

Answer 27

(d)

To determine

The change in the capacitance to make the power factor equal to 1.

(d)

Expert Solution

Answer to Problem 70P

$ΔC=1×10−4F$

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Cpf=1$ .The frequency is independent on the power factor and the inductance is remaining constant as per the conditions. So, $Cpf=1$ is establish as,

$CP f=1=1 ( 2πf )2LCP f=1=1 ( 2π×60Hz )2×0.05HCP f=1=1.4×10−4F$

The change in the capacitance for which the power factor is 1 is,

$ΔC=C−CP f=1ΔC=(2.4× 10 −4−1.4× 10 −4)FΔC=1×10−4F$

Conclusion:

The capacitance must change $ΔC=1×10−4F$

Answer 28

(d)

To determine

The change in the capacitance to make the power factor equal to 1.

Answer 29

(d)

Expert Solution

Answer to Problem 70P

$ΔC=1×10−4F$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given:

Rms voltage, $Vrms=120V$

Frequency, $f=60Hz$

Rms current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Cpf=1$ .The frequency is independent on the power factor and the inductance is remaining constant as per the conditions. So, $Cpf=1$ is establish as,

$CP f=1=1 ( 2πf )2LCP f=1=1 ( 2π×60Hz )2×0.05HCP f=1=1.4×10−4F$

The change in the capacitance for which the power factor is 1 is,

$ΔC=C−CP f=1ΔC=(2.4× 10 −4−1.4× 10 −4)FΔC=1×10−4F$

Conclusion:

The capacitance must change $ΔC=1×10−4F$

Answer 34

(e)

To determine

The change in inductance to make the power factor 1.

(e)

Expert Solution

Answer to Problem 70P

$ΔL=0.02H$

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH=0.05H$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Lpf=1$ . The frequency is independent of the power factor and the capacitance is remains constant as per the conditions. So, $Lpf=1$ is established as:

$LP f=1=1 ( 2πf )2CLP f=1=1 ( 2π×60Hz )2×2.4× 10 −4FLP f=1=0.03 H$

The change in the inductance for which the power factor is 1 is

$ΔL=L−LP f=1ΔL=(0.05−0.03) HΔL=0.02H$

Conclusion:

The change in inductance to make power factor 1 is $ΔL=0.02H$ .

Answer 35

(e)

To determine

The change in inductance to make the power factor 1.

Answer 36

(e)

Expert Solution

Answer to Problem 70P

$ΔL=0.02H$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given:

RMS voltage, $Vrms=120V$

Frequency, $f=60Hz$

RMS current, $Irms=11A$

Phase difference between the current and the voltage, $ϕ=45o$

Average power, $Pav=933.38W$

Inductance, $L=50mH=0.05H$

Capacitance, $C=2.4×10−4F$

Formula used:

Capacitive reactance, $XC=12πfC$

Frequency, $2πf=1LC$

Calculation:

The capacitance at power factor 1 is $Lpf=1$ . The frequency is independent of the power factor and the capacitance is remains constant as per the conditions. So, $Lpf=1$ is established as: