EBK COLLEGE PHYSICS, VOLUME 1
EBK COLLEGE PHYSICS, VOLUME 1
11th Edition
ISBN: 8220103599986
Author: Vuille
Publisher: Cengage Learning US
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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
Check Mark
To determine
The maximum value of current in the circuit.

Answer to Problem 30P

The maximum current in the circuit is 0.11A .

Explanation of Solution

Given Info: The maximum voltage of the AC source is 170V and has a frequency of 60Hz . A resistor of resistance 1.2 and a capacitor of capacitance 2.5μF 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 60Hz for f , 2.5μF for C to determine the capacitive reactance,

XC=12π(60Hz)(2.5μF)=1.1×103Ω

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

Z=R2+(XLXC)2

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

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

Z=(1.2)2+(01.1×103Ω)2=1.6×103Ω

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

Im=170V1.6×103Ω=0.11A

The maximum current in the circuit is 0.11A .

b)

Expert Solution
Check Mark
To determine
The maximum value of potential difference across the resistor and capacitor.

Answer to Problem 30P

The maximum voltage across the resistor is 1.3×102V and the capacitor is 1.2×102V

Explanation of Solution

Given Info: The maximum voltage of the AC source is 170V and has a frequency of 60Hz . A resistor of resistance 1.2 and a capacitor of capacitance 2.5μF 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 0.11A for Im and 1.1×103Ω for XC to determine the maximum potential difference across the resistor,

Vm,C=(0.11A)(1.1×103Ω)=1.2×102V

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

c)

Expert Solution
Check Mark
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 , 1.2×102V , 1.2×102V and 300μC respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is 170V and has a frequency of 60Hz . A resistor of resistance 1.2 and a capacitor of capacitance 2.5μF 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 0.11A for Im and 1.1×103Ω for XC to determine the maximum potential difference across the resistor,

Vm,C=(0.11A)(1.1×103Ω)=1.2×102V

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

ΔVsΔVRΔVC=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 , 1.2×102V for ΔVC to determine the voltage across the AC source,

ΔVs=0+1.2×102V=1.2×102V

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 1.2×102V .

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

q=(2.5μF)(1.2×102V)=(2.5×106F)(1.2×102V)=3.0×104C=300μC

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 , 1.2×102V , 1.2×102V and 300μC respectively.

d)

Expert Solution
Check Mark
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 1.3×102V , 0 , 1.3×102V , and 0 respectively.

Explanation of Solution

Given Info: The maximum voltage of the AC source is 170V and has a frequency of 60Hz . A resistor of resistance 1.2 and a capacitor of capacitance 2.5μF 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 0.11A for Im and 1.2 for R to determine the maximum potential difference across the resistor,

Vm,R=(0.11A)(1.2)=1.3×102V

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

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

ΔVs=1.3×102V+0=1.3×102V

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 2.5μF for C to determine the charge across the capacitor,

q=(2.5μF)(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 1.3×102V , 0 , 1.3×102V , and 0 respectively.

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Chapter 21 Solutions

EBK COLLEGE PHYSICS, VOLUME 1

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