Question

Solve the initial value problem

A 100-volt electromotive force is applied to an RC-series circuit in which the resistance is 200 ohms and the capacitance is 1024 farad. Find the charge q(t) on the capacitor if q(0) 5 0. Find the current i(t).

Short answer

Charge is$:\mathrm{q}\left(\mathrm{t}\right)=\left(\frac{1}{100}-\frac{1}{100}{\mathrm{e}}^{-50\mathrm{t}}\right)$

Charge is :$i\left(t\right)=\frac{1}{2}{e}^{-50t}$

Step-by-step solution

Definition of Series circuits

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

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

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$100$-volt electromotive force in a$RC$circuit in series with a resistance$\mathrm{R}=200$ohms and a capacitance$C={10}^{-4}$farad, and then we have the differential equation for Charge$\left(q\right)$as

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

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

Since we have $R=200\text{ohms},C={10}^{-4}\text{farad}$and$E=100$volt, then we have the differential equation shown in (1) as

$\begin{array}{c}200\frac{dq}{dt}+{10}^{4}q=100\times \frac{1}{200}\\ \frac{dq}{dt}+50q=0.5\end{array}$

$\frac{\mathrm{dq}}{\mathrm{dt}}=0.5-50\mathrm{q}$ --------(2)

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

Step 3: Separate the variables

$\begin{array}{c}\frac{dq}{0.5-50q}=dt\\ \int \frac{1}{0.5-50q}dq=\int dt\\ -\frac{1}{50}\int \frac{-50}{0.5-50q}dq=\int dt\\ -\frac{1}{50}\ln (0.5-50q)=t+c_1\end{array}$

$\begin{array}{c}\ln (0.5-50\mathrm{q})=-50\mathrm{t}+{\mathrm{c}}_2\\ {\mathrm{e}}^{\ln (0.5-50\mathrm{q})}={\mathrm{e}}^{(-50\mathrm{t}+{\mathrm{c}}_2)}\\ 0.5-50\mathrm{q}={\mathrm{e}}^{{\mathrm{c}}_2}{\mathrm{e}}^{-50\mathrm{t}}\\ 0.5-50\mathrm{q}={\mathrm{ce}}^{-50\mathrm{t}}\end{array}$

$50\mathrm{q}=0.5-{\mathrm{ce}}^{-50\mathrm{t}}$

Then we have

$\mathrm{q}\left(\mathrm{t}\right)=\frac{1}{100}-{\mathrm{ke}}^{-50\mathrm{t}}$--------(3)

After that, to find the value of constant $k$,we have to apply this point of condition$(\mathrm{q},\mathrm{t})=$ ( 0 coulomb , 0 minutes) into equation (3) as

$0\text{coulomb}=\frac{1}{100}-k{e}^0$

Then we have

$k=\frac{1}{100}$

Step 4: Substitution

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

$\mathrm{q}\left(\mathrm{t}\right)=\frac{1}{100}-\frac{1}{100}{\mathrm{e}}^{50\mathrm{t}}$--------(4)

is the charge on the capacitor of$\mathrm{RC}$series circuit at time $\mathrm{t}$.

Also, we have to obtain the current of this circuit as using equation (4) as

$\begin{array}{c}\mathrm{i}\left(\mathrm{t}\right)=\frac{\mathrm{dq}}{\mathrm{dt}}\\ =\frac{\mathrm{d}\left(\frac{1}{100}-\frac{1}{100}{\mathrm{e}}^{-50\mathrm{t}}\right)}{\mathrm{dt}}\\ =\frac{1}{2}{\mathrm{e}}^{-50\mathrm{t}}\end{array}$

is the current of $\mathrm{RC}$series circuit at time $t$.

Charge is $:\mathrm{q}\left(\mathrm{t}\right)=\left(\frac{1}{100}-\frac{1}{100}{\mathrm{e}}^{-50\mathrm{t}}\right)$

Charge is :$\mathrm{i}\left(\mathrm{t}\right)=\frac{1}{2}{\mathrm{e}}^{-50\mathrm{t}}$