Chapter 29, Problem 60PQ

Figure P29.60 shows a simple RC circuit with a 2.50-μF capacitor, a 3.50-MΩ resistor, a 9.00-V emf, and a switch.

What are

1. a. the charge on the capacitor,
2. b. the current in the resistor,
3. c. the rate at which the capacitor is storing energy, and
4. d. the rate at which the battery is delivering energy exactly 7.50 s alter the switch is closed?

To determine

Find the charge on the capacitor.

Answer to Problem 60PQ

The charge on the capacitor at $7.50\text{s}$ after closing the switch is $13.0\text{μC}$.

Explanation of Solution

Write the expression for the initial charge on a capacitor as.

  $q_o=C\epsilon$                                                                                                      (I)

Here, $q_o$ is the initial charge on the capacitor, $C$ is the capacitance in the circuit and $\epsilon$ is the emf of the battery.

Write the expression for the charge in a series RC circuit as.

  $q=q_o\left(1-e^{-\frac{t}{RC}}\right)$

Substitute $C\epsilon$ for $q_o$ in the above expression.

  $q=C\epsilon \left(1-e^{-\frac{t}{RC}}\right)$                                                                                      (II)

Here, $q$ is the charge on the capacitor after time $t$, $t$ is the time after which charge is to be determined and $R$ is the resistance of the circuit.

Conclusion:

Substitute $2.50\text{μf}$ for $C$, $3.50\text{MΩ}$ for $R$, $9.00\text{V}$ for $\epsilon$ and $7.50\text{s}$ for $t$ in equation (II).

  $\begin{array}{c}q=\left(2.50\text{μf}\right)\left(9.00\text{V}\right)\left(1-e^{-\frac{\left(7.50\text{s}\right)}{\left(3.50\text{MΩ}\right)\left(2.50\text{μf}\right)}}\right)\\ =\left(2.25\times {10}^{-5}\right)\left(1-e^{-0.857}\right)\text{C}\\ =\left(2.25\times {10}^{-5}\right)\left(0.575\right)\text{C}\\ =13.0\text{μC}\end{array}$

Thus, the charge on the capacitor at $7.50\text{s}$ after closing the switch is $13.0\text{μC}$.

To determine

Find the current in the circuit.

Answer to Problem 60PQ

The current in the circuit at $7.50\text{s}$ after closing the switch is $1.09\text{μA}$.

Explanation of Solution

Write the expression for the initial current in the circuit as.

  $I_o=\frac{\epsilon }{R}$                                                                                                        (III)

Here, $I_o$ is the initial current in the circuit.

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

  $I\left(t\right)=I_o{e}^{-\frac{t}{RC}}$

Substitute $\frac{\epsilon }{R}$ for $I_o$ in the above expression.

  $I\left(t\right)=\left(\frac{\epsilon }{R}\right)e^{-\frac{t}{RC}}$                                                                                         (IV)

Here, $I\left(t\right)$ is the current in the circuit at time $t$.

Conclusion:

Substitute $9.00\text{V}$ for $\epsilon$, $3.50\text{MΩ}$ for $R$ , $2.50\text{μf}$ for $C$ and $7.50\text{s}$ for $t$ in equation (IV).

  $\begin{array}{c}I\left(t\right)=\left(\frac{\left(9.00\text{V}\right)}{\left(3.50\text{MΩ}\right)}\right)e^{-\frac{\left(7.50\text{s}\right)}{\left(3.50\text{MΩ}\right)\left(2.50\text{μf}\right)}}\\ =\left(2.57\times {10}^{-6}\right)\left(e^{-0.857}\right)\text{A}\\ =\left(2.57\times {10}^{-6}\right)\left(0.424\right)\text{A}\\ \text{=1}\text{.09}\text{μA}\end{array}$

Thus, the current in the circuit at $7.50\text{s}$ after closing the switch is $1.09\text{μA}$.

To determine

Determine the rate at which the capacitor is storing energy.

Answer to Problem 60PQ

The rate of energy stored in the capacitor at $7.50\text{s}$ after closing the switch is $5.67\text{μW}$.

Explanation of Solution

Write the expression for the energy stored in a capacitor as.

  $U=\frac{1}{2}\frac{q^2}{C}$                                                                                                      (V)

Here, $U$ is the amount of energy stored in the capacitor.

Write the expression for the rate of storing energy in the capacitor as.

  $\begin{array}{c}\frac{dU}{dt}=\frac{1}{2C}\left(2q\frac{dq}{dt}\right)\\ =\frac{q}{C}\left(\frac{dq}{dt}\right)\end{array}$

Substitute $I\left(t\right)$ for $\frac{dq}{dt}$ in the above expression.

  $\frac{dU}{dt}=\frac{q}{C}I\left(t\right)$                                                                                              (VI)

Conclusion:

Substitute $13.0\text{μC}$ for $q$, $2.50\text{μF}$ for $C$ and $1.09\text{μA}$ for $I\left(t\right)$ in equation (VI).

  $\begin{array}{c}\frac{\text{d}U}{\text{dt}}\text{=}\frac{\left(\text{13}\text{.0}\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{\left(\text{2}\text{.50}\text{μF}\right)\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)}\left(\text{1}\text{.09}\text{μA}\right)\left(\frac{{10}^{-6}\text{A}}{1\text{μA}}\right)\\ \text{=5}{\text{.67×10}}^{-\text{6}}\text{W}\\ \text{=5}\text{.67}\text{μW}\end{array}$

Thus, the rate of energy stored in the capacitor at $7.50\text{s}$ after closing the switch is $5.67\text{μW}$.

To determine

The rate at which battery is delivering energy exactly $7.50\text{ s}$ after the switch is closed.

Answer to Problem 60PQ

The rate of energy delivered by battery at $7.50\text{s}$ after closing the switch is $9.82\text{μW}$.

Explanation of Solution

The rate of energy delivered by the battery is the power generated at the battery.

Write the expression for the rate of energy delivered by battery as.

  $P=\epsilon I\left(t\right)$                                                                                                   (VII)

Here, $P$ is the rate of energy delivered by the battery.

Conclusion:

Substitute $9.00\text{V}$ for $\epsilon$ and $1.09\text{μA}$ for $I\left(t\right)$ in equation (VII).

  $\begin{array}{c}P=\left(9.00\text{V}\right)\left(1.09\text{μA}\right)\\ =9.82\text{μW}\end{array}$

Thus, the rate of energy delivered by battery at $7.50\text{s}$ after closing the switch is $9.82\text{μW}$.