Question

A 200-volt electromotive force is applied to an RC-series circuit in which the resistance is 1000 ohms and the capacitance is 5 3 1026 farad. Find the charge q(t) on the capacitor if i(0) 5 0.4. Determine the charge and current at t 5 0.005 s. Determine the charge as $\mathit{t}\mathbf{}\mathbf{:}\mathbf{}\mathbf{\infty }$.

Short answer

The charge on the capacitor at time$t$ is : $q\left(t\right)=\frac{1}{1000}-\frac{1}{500}{e}^{-200t}$At $\text{t}=0.005$seconds we have$:q=2.64\times {10}^4$ coulomb and$i=14.72\times {10}^2\text{Amp}$ At $t\Rightarrow \infty$seconds we have$:q=1\times {10}^{-3}$ coulomb

Step-by-step solution

Definition of Series circuits

The linear differential equation for the current i(t),

$L\frac{di}{dt}+Ri=E\left(t\right)$

where L and R are known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system.

Differentiate the equation

We have a$200$-volt electromotive force in a$RC$circuit in series with a resistance$R=1000$ohms and a capacitance$C=5\times 10$farad, and then we have the differential equation for charge$\left(q\right)$as

$R\frac{dq}{dt}+\frac{1}{C}q=E\left(t\right)$-------- (1)

and we have to obtain the charge$q\left(t\right)$on the capacitor if we have the initial current as$i\left(0\right)=0.4$as the following:

Since we have$R=1000$ohms,$C=5\times {10}^{-6}$ farad and $E=200$volt, then we have the differential equation shown in (1) as

$\begin{array}{c}1000\frac{\mathrm{dq}}{\mathrm{dt}}+\frac{1}{5}\times {10}^6\mathrm{q}=200\\ \frac{\mathrm{dq}}{\mathrm{dt}}+200\mathrm{q}=0.2\end{array}$

$\frac{\mathrm{dq}}{\mathrm{dt}}=0.2-200\mathrm{q}$ -------- (2)

This equation is a first order linear and separable differential equation, then we can separate variables and then do integration as

Separate the variables 

$\begin{array}{c}\frac{\mathrm{dq}}{0.2-200\mathrm{q}}=\mathrm{dt}\\ \int \frac{1}{0.2-200\mathrm{q}}\mathrm{dq}=\int \mathrm{d}\mathrm{t}\\ -\frac{1}{200}\int \frac{-200}{0.2-200\mathrm{q}}\mathrm{dq}=\int \mathrm{d}\mathrm{t}\\ -\frac{1}{200}\ln (0.2-200\mathrm{q})=\mathrm{t}+{\mathrm{c}}_1\end{array}$

$\begin{array}{c}\ln (0.2-200\mathrm{q})=-200\mathrm{t}+{\mathrm{c}}_2\\ {\mathrm{e}}^{\ln (0.2-200\mathrm{q})}={\mathrm{e}}^{(-200\mathrm{t}+{\mathrm{c}}_2)}\\ 0.2-200\mathrm{q}={\mathrm{e}}^{{\mathrm{c}}_2}{\mathrm{e}}^{200\mathrm{t}}\\ 0.2-200\mathrm{q}={\mathrm{ce}}^{200\mathrm{t}}\end{array}$

$200\mathrm{q}=0.2-{\mathrm{ce}}^{-200\mathrm{t}}$

Then we have

$\mathrm{q}\left(\mathrm{t}\right)=\frac{1}{1000}-{\mathrm{ke}}^{-200\mathrm{t}}$ -------- (3)

After that, we have to obtain the current of this given circuit using equation (3) as

$\begin{array}{c}\mathrm{i}\left(\mathrm{t}\right)=\frac{\mathrm{dq}}{\mathrm{dt}}\\ =\frac{\mathrm{d}\left(\frac{1}{1000}-{\mathrm{ke}}^{-200\mathrm{t}}\right)}{\mathrm{dt}}\end{array}$

$=200k{e}^{-200t}$ -------- (4)

After that, to find the value of constant $k$, we have to apply this point of condition $(i,t)=(0.4\text{amp},0$seconds) into equation (3) as

$0.4\text{amp}=200k{e}^0$

$\begin{array}{rcl}k& =& \frac{0.4}{200}\\ & =& \frac{1}{500}\end{array}$

Substitution

After that, substitute with the value of into equation (3), then we have

$\mathrm{q}\left(\mathrm{t}\right)=\frac{1}{1000}-\frac{1}{500}{\mathrm{e}}^{-200\mathrm{t}}$-------- (5)

is the charge on the capacitor of $RC$series circuit at time$t$

Also, substitute with the value of into equation (4), then we have

$\mathrm{i}\left(\mathrm{t}\right)=\frac{2}{5}{\mathrm{e}}^{-200\mathrm{t}}$-------- (6)

is the current of $RC$series circuit at time $t$.

Now, we have to find the charge on the capacitor of $RC$series circuit at after $0.005$seconds by substituting with $t=\frac{1}{200}$second into equation (5), then we have

$\begin{array}{c}\mathrm{q}=\frac{1}{1000}-\frac{1}{500}{\mathrm{e}}^{\left(-200\left(\frac{1}{200}\right)\right)}\\ =\frac{1}{1000}-\frac{1}{500}{\mathrm{e}}^{-1}\\ =\frac{1}{1000}-\frac{1}{500}\times 0.367879\\ =1\times {10}^{-3}-7.36\times {10}^{-4}\end{array}$

$=2.64\times {10}^{-4}\text{coulomb}$

Also, we have to find the current of $RC$series circuit at after $0.0005$seconds by substituting with $t=\frac{1}{200}$second into equation (6), then we have

$\begin{array}{c}i=\frac{2}{5}{e}^{\left(-200\left(\frac{1}{200}\right)\right)}\\ =\frac{2}{5}{e}^{-1}\\ =\frac{2}{5}\times 0.367879\\ =14.72\times {10}^{-2}\text{Amp}\end{array}$

The charge on the capacitor at time$t$is$q\left(t\right)=\frac{1}{1000}-\frac{1}{500}{e}^{-200t}$ :At$\text{t}=0.005$ seconds we have$:q=2.64\times {10}^{4}coulomb$and$i=14.72\times {10}^2\text{Amp}$

At $t\Rightarrow \infty$seconds we have$:q=1\times {10}^{-3}$ coulomb.