Chapter 34, Problem 30SP

A series circuit consisting of an uncharged 2.0-µF capacitor and a 10-MΩ resistor is connected across a 100-V power source. What are the current in the circuit and the charge on the capacitor (a) after one time constant, and (b) when the capacitor has acquired 90 percent of its final charge?

To determine

The current in the circuit and charge across the capacitor in the RC circuit after one time constant when a series circuit consisting of uncharged $2.0\mu \text{F}$ capacitor and $10\text{MΩ}$ is connected to a $100\text{V}$ power supply.

Answer to Problem 30SP

Solution:

$3.7\text{ μA}$, $\text{0}\text{.13 mC}$

Explanation of Solution

Given data:

Capacitance of the capacitor is $\text{2}\text{.0 μF}$.

Resistance of the resistor is $10\text{ M}\Omega$.

The emf of the voltage source is $100\text{ V}$.

Formula used:

When a capacitor is connected to a battery, in the beginning the capacitor carries the current that keeps on decreasing till it is fully charged. Once the capacitor is fully charged the current through it is zero. This happens because a capacitor acts as a short circuit when it is uncharged and as an open circuit when it is fully charged.

The expression of current flowing through an RC circuit is

$I=I_0{e}^{-\left(\frac{t}{\tau }\right)}=I=I_0{e}^{-\left(\frac{t}{RC}\right)}$

Here, $I$ is the current at the time instant $t$, $I_0$ is the current when the capacitor is uncharged or just starting to charge, $R$ is the resistance, and $C$ is the capacitance.

The expression of the time constant in term of resistance and capacitance is

$\tau =RC$.

The maximum current in the circuit at starting is

$I_0=\frac{V_0}{R}$

Here, $V_0$ is the voltage of the voltage source.

The expression for instantaneous charge stored in the capacitor is

$Q=Q_0\left(1-e^{-\left(\frac{t}{\tau }\right)}\right)$

Here, $Q$ is the charge at the time instant $t$, and $Q_0$ is the charge when the capacitor voltage is equal to the source voltage.

The maximum charge stored in the capacitor.

$Q_0=C{V}_0$

Explanation:

Recall the expression of current flowing through an RC circuit.

$I=I_0{e}^{-\left(\frac{t}{\tau }\right)}$

Substitute $\tau$ for $t$, and $\frac{V_0}{R}$ for $I_0$.

$\begin{array}{c}I=\frac{V_0}{R}{e}^{-\left(\frac{\tau }{\tau }\right)}\\ =\frac{V_0}{R}{e}^{-1}\\ =\frac{V_0}{R}\left(\frac{1}{e}\right)\end{array}$

Substitute $100\text{ V}$ for $V_0$ and $10\text{ M}\Omega$ for $R$

$\begin{array}{c}I=\frac{100\text{ V}}{10\text{ M}\Omega }\left(\frac{1}{e}\right)\\ =3.7\text{ μA}\end{array}$

Recall the expression of instantaneous charge stored in the capacitor in RC circuit.

$Q=Q_0\left(1-e^{-\left(\frac{t}{\tau }\right)}\right)$

Substitute $\tau$ for $t$, and $C{V}_0$ for $Q_0$.

$\begin{array}{c}Q=C{V}_0\left(1-e^{-\left(\frac{\tau }{\tau }\right)}\right)\\ =C{V}_0\left(1-e^{-1}\right)\\ =C{V}_0\left(1-\frac{1}{e}\right)\end{array}$

Substitute $100\text{ V}$ for $V_0$, and $2.0\text{ μF}$ for $C$.

$\begin{array}{c}Q=\left(2.0\text{ μF}\right)\left(100\text{ V}\right)\left(1-\frac{1}{e}\right)\\ =0.13\text{ mC}\end{array}$

Conclusion:

The current in the RC circuit is $3.7\text{ μA}$ and charge across the capacitor after one-time constant is $\text{0}\text{.13 mC}$.

To determine

The current in the RC circuit and charge across the capacitor when the capacitor acquires 90 percent of its final chargewhen a series circuit consisting of uncharged $2.0\mu \text{F}$ capacitor and $10\text{MΩ}$ is connected to a $100\text{V}$ power supply.

Answer to Problem 30SP

Solution:

$1.0\text{ μA}$, $0.18\text{ mC}$

Explanation of Solution

Given data:

Capacitance of the capacitor is $\text{2}\text{.0 μF}$.

Resistance of the resistor is $10\text{ M}\Omega$.

The emf of the voltage source is $100\text{ V}$.

Formula used:

When a capacitor is connected to a battery, in the beginning the capacitor carries the current that keeps on decreasing till it is fully charged. Once the capacitor is fully charged the current through it is zero. This happens because a capacitor acts as a short circuit when it is uncharged and as an open circuit when it is fully charged.

The expression for instantaneous charge stored in the capacitor is

$Q=Q_0\left(1-e^{-\left(\frac{t}{\tau }\right)}\right)$

Here, $Q$ is the charge at the time instant $t$, and $Q_0$ is the charge when the capacitor voltage is equal to the source voltage.

The maximum charge stored in the capacitor is

$Q_0=C{V}_0$

Here, $V_0$ is the voltage of the voltage source.

The expression of current flowing through an RC circuit when capacitor is charging is

$I=I_0{e}^{-\left(\frac{t}{\tau }\right)}$

Here, $I$ is the current at the time instant $t$, $I_0$ is the current when the capacitor is uncharged or just started to charge, $R$ is the resistance, and $C$ is the capacitance.

The expression of the time constant in term of resistance and capacitance is

$\tau =RC$.

The maximum current in the circuit at starting is

$I_0=\frac{V_0}{R}$

Explanation:

Recall the expression for instantaneous charge stored in the capacitor.

$Q=Q_0\left(1-e^{-\left(\frac{t}{\tau }\right)}\right)$

The capacitor stored $90\text{ \%}$ of the final charge.

Substitute $\tau$ for $RC$, and $0.90\left(Q_0\right)$ for $Q$

$\begin{array}{c}0.90\left(Q_0\right)=Q_0\left(1-e^{-\left(\frac{t}{RC}\right)}\right)\\ \frac{9}{10}=\left(1-e^{-\left(\frac{t}{RC}\right)}\right)\\ e^{-\left(\frac{t}{RC}\right)}=\left(1-\frac{9}{10}\right)\\ =\frac{1}{10}\end{array}$

Take logarithm on both sides

$\begin{array}{c}\ln \left(e^{-\left(\frac{t}{RC}\right)}\right)=\ln \left(\frac{1}{10}\right)\\ -\left(\frac{t}{RC}\right)=-2.3\\ t=2.3\left(RC\right)\end{array}$

Substitute $10\text{ M}\Omega$ for $R$ and $\text{2}\text{.0 μF}$ for $C$

$\begin{array}{c}t=2.3\left(10\text{ M}\Omega \right)\left(\text{2}\text{.0 μF}\right)\\ =46\text{ sec}\end{array}$

Recall the expression of current flowing through an RC circuit when capacitor is charging.

$I=I_0{e}^{-\left(\frac{t}{\tau }\right)}$

Substitute $\tau$ for $RC$

$I=I_0{e}^{-\left(\frac{t}{RC}\right)}$

Substitute $\frac{V_0}{R}$ for $I_0$

$I=\frac{V_0}{R}{e}^{-\left(\frac{t}{RC}\right)}$

Substitute $10\text{ M}\Omega$ for $R$, $100\text{ V}$ for $V_0$, $46\text{ sec}$ for $t$, and $\text{2}\text{.0 μF}$ for $C$

$\begin{array}{c}I=\frac{100\text{ V}}{10\text{ M}\Omega }{e}^{-\left(\frac{46\text{ sec}}{\left(10\text{ M}\Omega \right)\left(\text{2}\text{.0 μF}\right)}\right)}\\ =1.0\text{}\text{ μA}\end{array}$

Recall the expression of the maximum charge stored in the capacitor.

$Q_0=C{V}_0$

Substitute $100\text{ V}$ for $V_0$, and $2.0\text{ μF}$ for $C$

$\begin{array}{c}{Q}_0=\left(2.0\text{ μF}\right)\left(100\text{ V}\right)\\ =2\times {10}^{-4}\text{ C}\end{array}$

The capacitor acquired 90 percent of its final charge

$\begin{array}{c}Q=90\%\text{ of }{Q}_0\\ =0.90Q_0\end{array}$

Substitute $2\times {10}^{-4}\text{ C}$ for $Q_0$

$\begin{array}{c}Q=0.90\left(2\times {10}^{-4}\text{ C}\right)\\ =0.18\text{ mC}\end{array}$

Conclusion:

The current in the RC circuit is $1.0\text{ μA}$ and charge across the capacitor is $0.18\text{ mC}$ when the capacitor acquires 90% of its final charge.