The charge on the capacitor plates.

Question

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Answer 1

Question

Chapter 17, Problem 77E

(a)

To determine

The charge on the capacitor plates.

(a)

Expert Solution

Answer to Problem 77E

The charge on the capacitor plates is $0.1 C_$ .

Explanation of Solution

Given that the capacitance is $100 μF$ , and the potential difference is $1000 V$ .

Write the expression for charge on the capacitor plates.

$Q=CV$ (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $100 μF$ for $C$ , and $1000 V$ for $V$ in equation (I) to find $Q$ .

$Q=(100 μF)(1000 V)=(100 μF×1 F1×106 μF)(1000 V)=0.1 C$

Therefore, the charge on the capacitor plates is $0.1 C_$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 77E

The resistance of the circuit is $10 ohm_$ .

Explanation of Solution

Given that the time is $0.001 s$ , the remaining charge is $0.37$ times the initial charge, and the capacitance is $100 μF$ .

Write the expression for the charge in the capacitor in an RC circuit under charging.

$q(t)=qfe−t/RC$ (II)

Here, $q(t)$ is the charge at time $t$ , $R$ is the resistance, and $C$ is the capacitance, $qf$ is the charge when capacitor is fully charged (here the initial charge).

Since the charge remaining is $0.37$ of the initial charge, in equation (II), $q(t)$ can be replaced by $0.37qf$ .

$0.37qf=qfe−t/RC0.37=e−t/RC$ (III)

Solve equation (III) for $R$

$ln0.37=−tRCR=−tCln0.37$ (IV)

Conclusion:

Substitute $0.001 s$ for $t$ , and $100 μF$ for $C$ in equation (IV) to find $R$ .

$R=−(0.001 s)(100 μF)ln0.63=−(0.001 s)(100 μF×1 F1×106 μF)ln0.37=10 ohm$

Therefore, the resistance of the circuit is $10 ohm_$ .

(c)

To determine

The current after $0.001 s$ .

(c)

Expert Solution

Answer to Problem 77E

The current after $0.001 s$ is $37 A_$ .

Explanation of Solution

Write the expression for the current in the series RC circuit.

$i(t)=εRe−t/RC$ (V)

Here, $i(t)$ is the current at time $t$ , $ε$ is the potential difference.

Conclusion:

Substitute $10 ohm$ for $R$ , $100 μF$ for $C$ , $1000 V$ for $ε$ , and $0.001 s$ for $t$ in equation (V) to find $i$ .

$i(t)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF×1 F1×106 μF)=37 A$

Therefore, the current after $0.001 s$ is $37 A_$ .

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Answer 2

Question

Chapter 17, Problem 77E

(a)

To determine

The charge on the capacitor plates.

(a)

Expert Solution

Answer to Problem 77E

The charge on the capacitor plates is $0.1 C_$ .

Explanation of Solution

Given that the capacitance is $100 μF$ , and the potential difference is $1000 V$ .

Write the expression for charge on the capacitor plates.

$Q=CV$ (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $100 μF$ for $C$ , and $1000 V$ for $V$ in equation (I) to find $Q$ .

$Q=(100 μF)(1000 V)=(100 μF×1 F1×106 μF)(1000 V)=0.1 C$

Therefore, the charge on the capacitor plates is $0.1 C_$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 77E

The resistance of the circuit is $10 ohm_$ .

Explanation of Solution

Given that the time is $0.001 s$ , the remaining charge is $0.37$ times the initial charge, and the capacitance is $100 μF$ .

Write the expression for the charge in the capacitor in an RC circuit under charging.

$q(t)=qfe−t/RC$ (II)

Here, $q(t)$ is the charge at time $t$ , $R$ is the resistance, and $C$ is the capacitance, $qf$ is the charge when capacitor is fully charged (here the initial charge).

Since the charge remaining is $0.37$ of the initial charge, in equation (II), $q(t)$ can be replaced by $0.37qf$ .

$0.37qf=qfe−t/RC0.37=e−t/RC$ (III)

Solve equation (III) for $R$

$ln0.37=−tRCR=−tCln0.37$ (IV)

Conclusion:

Substitute $0.001 s$ for $t$ , and $100 μF$ for $C$ in equation (IV) to find $R$ .

$R=−(0.001 s)(100 μF)ln0.63=−(0.001 s)(100 μF×1 F1×106 μF)ln0.37=10 ohm$

Therefore, the resistance of the circuit is $10 ohm_$ .

(c)

To determine

The current after $0.001 s$ .

(c)

Expert Solution

Answer to Problem 77E

The current after $0.001 s$ is $37 A_$ .

Explanation of Solution

Write the expression for the current in the series RC circuit.

$i(t)=εRe−t/RC$ (V)

Here, $i(t)$ is the current at time $t$ , $ε$ is the potential difference.

Conclusion:

Substitute $10 ohm$ for $R$ , $100 μF$ for $C$ , $1000 V$ for $ε$ , and $0.001 s$ for $t$ in equation (V) to find $i$ .

$i(t)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF×1 F1×106 μF)=37 A$

Therefore, the current after $0.001 s$ is $37 A_$ .

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Answer 3

Question

Chapter 17, Problem 77E

(a)

To determine

The charge on the capacitor plates.

(a)

Expert Solution

Answer to Problem 77E

The charge on the capacitor plates is $0.1 C_$ .

Explanation of Solution

Given that the capacitance is $100 μF$ , and the potential difference is $1000 V$ .

Write the expression for charge on the capacitor plates.

$Q=CV$ (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $100 μF$ for $C$ , and $1000 V$ for $V$ in equation (I) to find $Q$ .

$Q=(100 μF)(1000 V)=(100 μF×1 F1×106 μF)(1000 V)=0.1 C$

Therefore, the charge on the capacitor plates is $0.1 C_$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 77E

The resistance of the circuit is $10 ohm_$ .

Explanation of Solution

Given that the time is $0.001 s$ , the remaining charge is $0.37$ times the initial charge, and the capacitance is $100 μF$ .

Write the expression for the charge in the capacitor in an RC circuit under charging.

$q(t)=qfe−t/RC$ (II)

Here, $q(t)$ is the charge at time $t$ , $R$ is the resistance, and $C$ is the capacitance, $qf$ is the charge when capacitor is fully charged (here the initial charge).

Since the charge remaining is $0.37$ of the initial charge, in equation (II), $q(t)$ can be replaced by $0.37qf$ .

$0.37qf=qfe−t/RC0.37=e−t/RC$ (III)

Solve equation (III) for $R$

$ln0.37=−tRCR=−tCln0.37$ (IV)

Conclusion:

Substitute $0.001 s$ for $t$ , and $100 μF$ for $C$ in equation (IV) to find $R$ .

$R=−(0.001 s)(100 μF)ln0.63=−(0.001 s)(100 μF×1 F1×106 μF)ln0.37=10 ohm$

Therefore, the resistance of the circuit is $10 ohm_$ .

(c)

To determine

The current after $0.001 s$ .

(c)

Expert Solution

Answer to Problem 77E

The current after $0.001 s$ is $37 A_$ .

Explanation of Solution

Write the expression for the current in the series RC circuit.

$i(t)=εRe−t/RC$ (V)

Here, $i(t)$ is the current at time $t$ , $ε$ is the potential difference.

Conclusion:

Substitute $10 ohm$ for $R$ , $100 μF$ for $C$ , $1000 V$ for $ε$ , and $0.001 s$ for $t$ in equation (V) to find $i$ .

$i(t)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF×1 F1×106 μF)=37 A$

Therefore, the current after $0.001 s$ is $37 A_$ .

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Answer 4

Question

Chapter 17, Problem 77E

(a)

To determine

The charge on the capacitor plates.

(a)

Expert Solution

Answer to Problem 77E

The charge on the capacitor plates is $0.1 C_$ .

Explanation of Solution

Given that the capacitance is $100 μF$ , and the potential difference is $1000 V$ .

Write the expression for charge on the capacitor plates.

$Q=CV$ (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $100 μF$ for $C$ , and $1000 V$ for $V$ in equation (I) to find $Q$ .

$Q=(100 μF)(1000 V)=(100 μF×1 F1×106 μF)(1000 V)=0.1 C$

Therefore, the charge on the capacitor plates is $0.1 C_$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 77E

The resistance of the circuit is $10 ohm_$ .

Explanation of Solution

Given that the time is $0.001 s$ , the remaining charge is $0.37$ times the initial charge, and the capacitance is $100 μF$ .

Write the expression for the charge in the capacitor in an RC circuit under charging.

$q(t)=qfe−t/RC$ (II)

Here, $q(t)$ is the charge at time $t$ , $R$ is the resistance, and $C$ is the capacitance, $qf$ is the charge when capacitor is fully charged (here the initial charge).

Since the charge remaining is $0.37$ of the initial charge, in equation (II), $q(t)$ can be replaced by $0.37qf$ .

$0.37qf=qfe−t/RC0.37=e−t/RC$ (III)

Solve equation (III) for $R$

$ln0.37=−tRCR=−tCln0.37$ (IV)

Conclusion:

Substitute $0.001 s$ for $t$ , and $100 μF$ for $C$ in equation (IV) to find $R$ .

$R=−(0.001 s)(100 μF)ln0.63=−(0.001 s)(100 μF×1 F1×106 μF)ln0.37=10 ohm$

Therefore, the resistance of the circuit is $10 ohm_$ .

(c)

To determine

The current after $0.001 s$ .

(c)

Expert Solution

Answer to Problem 77E

The current after $0.001 s$ is $37 A_$ .

Explanation of Solution

Write the expression for the current in the series RC circuit.

$i(t)=εRe−t/RC$ (V)

Here, $i(t)$ is the current at time $t$ , $ε$ is the potential difference.

Conclusion:

Substitute $10 ohm$ for $R$ , $100 μF$ for $C$ , $1000 V$ for $ε$ , and $0.001 s$ for $t$ in equation (V) to find $i$ .

$i(t)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF×1 F1×106 μF)=37 A$

Therefore, the current after $0.001 s$ is $37 A_$ .

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Answer 5

Chapter 17, Problem 77E

(a)

To determine

The charge on the capacitor plates.

(a)

Expert Solution

Answer to Problem 77E

The charge on the capacitor plates is $0.1 C_$ .

Explanation of Solution

Given that the capacitance is $100 μF$ , and the potential difference is $1000 V$ .

Write the expression for charge on the capacitor plates.

$Q=CV$ (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $100 μF$ for $C$ , and $1000 V$ for $V$ in equation (I) to find $Q$ .

$Q=(100 μF)(1000 V)=(100 μF×1 F1×106 μF)(1000 V)=0.1 C$

Therefore, the charge on the capacitor plates is $0.1 C_$ .

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 77E

The resistance of the circuit is $10 ohm_$ .

Explanation of Solution

Given that the time is $0.001 s$ , the remaining charge is $0.37$ times the initial charge, and the capacitance is $100 μF$ .

Write the expression for the charge in the capacitor in an RC circuit under charging.

$q(t)=qfe−t/RC$ (II)

Here, $q(t)$ is the charge at time $t$ , $R$ is the resistance, and $C$ is the capacitance, $qf$ is the charge when capacitor is fully charged (here the initial charge).

Since the charge remaining is $0.37$ of the initial charge, in equation (II), $q(t)$ can be replaced by $0.37qf$ .

$0.37qf=qfe−t/RC0.37=e−t/RC$ (III)

Solve equation (III) for $R$

$ln0.37=−tRCR=−tCln0.37$ (IV)

Conclusion:

Substitute $0.001 s$ for $t$ , and $100 μF$ for $C$ in equation (IV) to find $R$ .

$R=−(0.001 s)(100 μF)ln0.63=−(0.001 s)(100 μF×1 F1×106 μF)ln0.37=10 ohm$

Therefore, the resistance of the circuit is $10 ohm_$ .

(c)

To determine

The current after $0.001 s$ .

(c)

Expert Solution

Answer to Problem 77E

The current after $0.001 s$ is $37 A_$ .

Explanation of Solution

Write the expression for the current in the series RC circuit.

$i(t)=εRe−t/RC$ (V)

Here, $i(t)$ is the current at time $t$ , $ε$ is the potential difference.

Conclusion:

Substitute $10 ohm$ for $R$ , $100 μF$ for $C$ , $1000 V$ for $ε$ , and $0.001 s$ for $t$ in equation (V) to find $i$ .

$i(t)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF×1 F1×106 μF)=37 A$

Therefore, the current after $0.001 s$ is $37 A_$ .

Answer 6

Chapter 17, Problem 77E

(a)

To determine

The charge on the capacitor plates.

(a)

Expert Solution

Answer to Problem 77E

The charge on the capacitor plates is $0.1 C_$ .

Explanation of Solution

Given that the capacitance is $100 μF$ , and the potential difference is $1000 V$ .

Write the expression for charge on the capacitor plates.

$Q=CV$ (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $100 μF$ for $C$ , and $1000 V$ for $V$ in equation (I) to find $Q$ .

$Q=(100 μF)(1000 V)=(100 μF×1 F1×106 μF)(1000 V)=0.1 C$

Therefore, the charge on the capacitor plates is $0.1 C_$ .

Answer 7

Chapter 17, Problem 77E

(a)

To determine

The charge on the capacitor plates.

Answer 8

(a)

Expert Solution

Answer to Problem 77E

The charge on the capacitor plates is $0.1 C_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given that the capacitance is $100 μF$ , and the potential difference is $1000 V$ .

Write the expression for charge on the capacitor plates.

$Q=CV$ (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $100 μF$ for $C$ , and $1000 V$ for $V$ in equation (I) to find $Q$ .

$Q=(100 μF)(1000 V)=(100 μF×1 F1×106 μF)(1000 V)=0.1 C$

Therefore, the charge on the capacitor plates is $0.1 C_$ .

Answer 13

(b)

To determine

The resistance of the circuit.

(b)

Expert Solution

Answer to Problem 77E

The resistance of the circuit is $10 ohm_$ .

Explanation of Solution

Given that the time is $0.001 s$ , the remaining charge is $0.37$ times the initial charge, and the capacitance is $100 μF$ .

Write the expression for the charge in the capacitor in an RC circuit under charging.

$q(t)=qfe−t/RC$ (II)

Here, $q(t)$ is the charge at time $t$ , $R$ is the resistance, and $C$ is the capacitance, $qf$ is the charge when capacitor is fully charged (here the initial charge).

Since the charge remaining is $0.37$ of the initial charge, in equation (II), $q(t)$ can be replaced by $0.37qf$ .

$0.37qf=qfe−t/RC0.37=e−t/RC$ (III)

Solve equation (III) for $R$

$ln0.37=−tRCR=−tCln0.37$ (IV)

Conclusion:

Substitute $0.001 s$ for $t$ , and $100 μF$ for $C$ in equation (IV) to find $R$ .

$R=−(0.001 s)(100 μF)ln0.63=−(0.001 s)(100 μF×1 F1×106 μF)ln0.37=10 ohm$

Therefore, the resistance of the circuit is $10 ohm_$ .

Answer 14

(b)

To determine

The resistance of the circuit.

Answer 15

(b)

Expert Solution

Answer to Problem 77E

The resistance of the circuit is $10 ohm_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given that the time is $0.001 s$ , the remaining charge is $0.37$ times the initial charge, and the capacitance is $100 μF$ .

Write the expression for the charge in the capacitor in an RC circuit under charging.

$q(t)=qfe−t/RC$ (II)

Here, $q(t)$ is the charge at time $t$ , $R$ is the resistance, and $C$ is the capacitance, $qf$ is the charge when capacitor is fully charged (here the initial charge).

Since the charge remaining is $0.37$ of the initial charge, in equation (II), $q(t)$ can be replaced by $0.37qf$ .

$0.37qf=qfe−t/RC0.37=e−t/RC$ (III)

Solve equation (III) for $R$

$ln0.37=−tRCR=−tCln0.37$ (IV)

Conclusion:

Substitute $0.001 s$ for $t$ , and $100 μF$ for $C$ in equation (IV) to find $R$ .

$R=−(0.001 s)(100 μF)ln0.63=−(0.001 s)(100 μF×1 F1×106 μF)ln0.37=10 ohm$

Therefore, the resistance of the circuit is $10 ohm_$ .

Answer 20

(c)

To determine

The current after $0.001 s$ .

(c)

Expert Solution

Answer to Problem 77E

The current after $0.001 s$ is $37 A_$ .

Explanation of Solution

Write the expression for the current in the series RC circuit.

$i(t)=εRe−t/RC$ (V)

Here, $i(t)$ is the current at time $t$ , $ε$ is the potential difference.

Conclusion:

Substitute $10 ohm$ for $R$ , $100 μF$ for $C$ , $1000 V$ for $ε$ , and $0.001 s$ for $t$ in equation (V) to find $i$ .

$i(t)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF×1 F1×106 μF)=37 A$

Therefore, the current after $0.001 s$ is $37 A_$ .

Answer 21

(c)

To determine

The current after $0.001 s$ .

Answer 22

(c)

Expert Solution

Answer to Problem 77E

The current after $0.001 s$ is $37 A_$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for the current in the series RC circuit.

$i(t)=εRe−t/RC$ (V)

Here, $i(t)$ is the current at time $t$ , $ε$ is the potential difference.

Conclusion:

Substitute $10 ohm$ for $R$ , $100 μF$ for $C$ , $1000 V$ for $ε$ , and $0.001 s$ for $t$ in equation (V) to find $i$ .

$i(t)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF)=1000 V10 ohme−0.001 s/(10 ohm)(100 μF×1 F1×106 μF)=37 A$

Therefore, the current after $0.001 s$ is $37 A_$ .

Answer 27

Ch. 17 - Prob. 1RQ Ch. 17 - Prob. 2RQ Ch. 17 - Prob. 3RQ Ch. 17 - Prob. 4RQ Ch. 17 - Prob. 5RQ Ch. 17 - Prob. 6RQ Ch. 17 - Prob. 7RQ Ch. 17 - Prob. 8RQ Ch. 17 - Prob. 9RQ Ch. 17 - Prob. 10RQ

The charge on the capacitor plates.

Videos

The charge on the capacitor plates.

Answer to Problem 77E

Explanation of Solution

The resistance of the circuit.

Answer to Problem 77E

Explanation of Solution

The current after $0.001 s$ .

Answer to Problem 77E

Explanation of Solution

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Chapter 17 Solutions

The charge on the capacitor plates.

Videos

The charge on the capacitor plates.

Answer to Problem 77E

Explanation of Solution

The resistance of the circuit.

Answer to Problem 77E

Explanation of Solution

The current after 0.001 s<m:math><m:mrow><m:mn>0.001</m:mn><m:mtext> </m:mtext><m:mtext>s</m:mtext></m:mrow></m:math>.

Answer to Problem 77E

Explanation of Solution

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Chapter 17 Solutions

The current after $0.001 s$ .