Chapter 8, Problem 1E

A source-free RC circuit has R = 4 kΩ and C = 22 μF, and with the knowledge that v(0) = 5 V, (a) write an expression for v(t) valid for t > 0; (b) compute v(t) at t = 0, t = 50 ms, and t = 500 ms; and (c) calculate the energy stored in the capacitor at t = 0, t = 50 ms, and t = 500 ms.

To determine

Write an expression for $v\left(t\right)$ valid for $t>0$.

Answer to Problem 1E

The expression for $v\left(t\right)$ valid for $t>0$ is $5e^{\frac{-t}{88\text{ ms}}}\text{V}$.

Explanation of Solution

Given Data:

The initial value of the voltage across the capacitor is $5\text{ V}$,

The resistance of $RC$ circuit is $4\text{ k}\Omega$ and

The capacitance of $RC$ circuit is $22\mu \text{F}$.

Formula used:

The expression for the time constant for $RC$ circuit is as follows,

$\tau =R_{eq}C$ (1)

Here,

$\tau$ is the time constant,

$R_{eq}$ is the equivalent resistance across the capacitor and

$C$ is the capacitance.

The expression for the voltage across the capacitor for $t>0$ is as follows,

$v(t)=v(0)e^{\frac{-t}{\tau }}$ (2)

Here,

$v(t)$ is the voltage across capacitor at $t>0$,

$v(0)$ is the initial value of the voltage across the capacitor and

$t$ is the time.

Calculation:

Substitute $4\text{ k}\Omega$ for $R_{eq}$ and $22\mu \text{F}$ for $C$ in equation (1),

$\begin{array}{c}\tau =\left(4\text{ k}\Omega \right)\left(22\mu \text{F}\right)\\ =\left(4000\Omega \right)\left(22\mu \text{F}\right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =\left(4000\Omega \right)\left(22\times {10}^{-6}\text{ F}\right)\left\{\because 1\mu \text{F}={10}^{-6}\text{ F}\right\}\\ =0.088\text{ s}\end{array}$

Substitute $5\text{ V}$ for $v(0)$ and $0.088\text{ s}$ for $\tau$ in equation (2),

$v(t)=\left(5\text{V}\right)e^{\frac{-t}{0.088\text{ s}}}$

$v(t)=5e^{\frac{-t}{88\text{ ms}}}\text{ V }\left\{\because \text{ 1 s}={10}^3\text{ ms}\right\}$ (3)

Conclusion:

Thus, the expression for $v\left(t\right)$ valid for $t>0$ is $5e^{\frac{-t}{88\text{ ms}}}\text{ V}$.

To determine

Find $v\left(t\right)$ at $t=0\text{ s}$, $t=50\text{ ms}$ and $t=500\text{ ms}$.

Answer to Problem 1E

The value of the voltage $v\left(t\right)$ at $t=0\text{ s}$ is $5\text{ V}$, at $t=50\text{ ms}$ is $2.8328\text{ V}$ and at $t=500\text{ ms}$ is $17.0\text{mV}$.

Explanation of Solution

Given Data:

The time at which voltage has to be calculated is $0\text{ s}$, $50\text{ ms}$ and $500\text{ ms}$.

Calculation:

Substitute $0\text{ s}$ for $t$ in equation (3),

$\begin{array}{c}v(t)=\left(5\text{V}\right)e^{\frac{-0\text{ s}}{88\text{ ms}}}\\ =\left(5\text{V}\right)\left(1\right)\\ =5\text{V}\end{array}$

Substitute $50\text{ ms}$ for $t$ in the equation (3),

$\begin{array}{c}v(t)=\left(5\text{V}\right)e^{\frac{-50\text{ ms}}{88\text{ ms}}}\\ =\left(5\text{V}\right)e^{\frac{-50}{88}}\\ =\left(5\text{V}\right)\left(0.56655\right)\\ =2.8328\text{ V}\end{array}$

Substitute $500\text{ ms}$ for $t$ in equation (3),

$\begin{array}{c}v(t)=\left(5\text{V}\right)e^{\frac{-500\text{ ms}}{88\text{ ms}}}\\ =\left(5\text{V}\right)e^{\frac{-500}{88}}\\ =\left(5\text{V}\right)\left(3.40736\right)\\ =0.0170\text{ V}\end{array}$

$=17.0\text{mV }\left\{\because 1\text{ V}={10}^3\text{ mV}\right\}$

Conclusion:

Thus, the value of the voltage $v\left(t\right)$ at $t=0\text{ s}$ is $5\text{ V}$, at $t=50\text{ ms}$  is $2.8328\text{ V}$ and at $t=500\text{ ms}$ is $17.0\text{mV}$.

To determine

Find the value of energy stored in the capacitor for different time periods.

Answer to Problem 1E

The energy stored in the capacitor at $t=0\text{ s}$ is $275\mu \text{J}$, at $t=50\text{ ms}$ is $88.27\mu \text{J}$ and at $t=500\text{ ms}$ is $3.179\text{ nJ}$.

Explanation of Solution

Given Data:

The time is $0\text{ s}$, $50\text{ ms}$ and $500\text{ ms}$.

Formula used:

The expression for the energy stored in the capacitor is as follows.

$w=\frac{1}{2}C{\left(v(t)\right)}^2$ (4)

Here,

$w$ is the energy stored in the capacitor,

$C$ is the capacitance and

$v(t)$ is the voltage across capacitor at the given time,

Calculation:

Substitute $5\text{ V}$ for $v\left(t\right)$ and $22\mu \text{F}$ for $C$ in equation (4),

$\begin{array}{c}w=\frac{1}{2}\left(22\mu \text{F}\right){\left(5\text{ V}\right)}^2\\ =\frac{1}{2}\left(22\times {10}^{-6}\text{ F}\right)\left(25\text{ V}\right)\left\{\because 1\mu \text{F}={10}^{-6}\text{ F}\right\}\\ =\frac{1}{2}\left(5.5\times {10}^{-4}\right)\text{ J}\\ =2.75\times {10}^{-4}\text{ J}\end{array}$

$\begin{array}{c}=275\times {10}^{-6}\text{ J}\\ \text{=}275\mu \text{J }\left\{\because {10}^{-6}\text{ J}=1\mu \text{J}\right\}\end{array}$

Substitute $2.8328\text{ V}$ for $v\left(t\right)$ and $22\mu \text{F}$ for $C$ in equation (4),

$\begin{array}{c}w=\frac{1}{2}\left(22\mu \text{F}\right){\left(2.8328\text{ V}\right)}^2\\ =\frac{1}{2}\left(22\times {10}^{-6}\text{ F}\right)\left(8.0247\text{ V}\right)\left\{\because 1\mu \text{F}={10}^{-6}\text{ F}\right\}\\ =\frac{1}{2}\left(1.7655\times {10}^{-4}\right)\text{ J}\\ =8.827\times {10}^{-5}\text{ J}\end{array}$

$\begin{array}{c}=8.827\times {10}^{-5}\text{ J}\\ =88.27\mu \text{J }\left\{\because {10}^{-6}\text{ J}=1\mu \text{J}\right\}\end{array}$

Substitute $0.0170\text{ V}$ for $v\left(t\right)$ and $22\mu \text{F}$ for $C$ in equation (4),

$\begin{array}{c}w=\frac{1}{2}\left(22\mu \text{F}\right){\left(0.0170\text{ V}\right)}^2\\ =\frac{1}{2}\left(22\times {10}^{-6}\text{ F}\right)\left(2.89\times {10}^{-4}\text{ V}\right)\left\{\because 1\mu \text{F}={10}^{-6}\text{ F}\right\}\\ =\frac{1}{2}\left(6.358\times {10}^{-9}\right)\text{ J}\\ =3.179\times {10}^{-9}\text{ J}\end{array}$

$=3.179\text{ nJ }\left\{\because {10}^{-9}\text{ J}=1\text{ nJ}\right\}$

Conclusion:

Thus, the energy stored in the capacitor at $t=0\text{ s}$ is $275\mu \text{J}$, at $t=50\text{ ms}$ is $88.27\mu \text{J}$ and at $t=500\text{ ms}$ is $3.179\text{ nJ}$.