Question

A 30-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1henry and the resistance is 50ohms. Find the current$\mathbf{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$if. Determine the current as$\mathbf{t}\mathbf{\to }\mathbf{\infty }$.

Short answer

The current is $\text{i(t) =}\frac{\text{3}}{\text{5}}\text{-}\frac{\text{3}}{\text{5}}{\text{e}}^{\text{- 500t}}$. And the current as$\mathrm{t}=\infty$ is$\frac{3}{5}$ amp.

Step-by-step solution

Define series circuit

For a series circuit containing only a resistor and an inductor, Kirchhoff’s second law states that the sum of the voltage drop across the inductor $\mathbf{\left(}\mathbf{L}\mathbf{\right(}\mathbf{di}\mathbf{/}\mathbf{dt}\mathbf{\left)}\mathbf{\right)}$and the voltage drop across the resistor $\mathbf{\left(}\mathbf{iR}\mathbf{\right)}$is the same as the impressed voltage $\mathbf{\left(}\mathbf{E}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right)}$on the circuit.

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

Solve for first order series circuit equation.

Let the differential equation for current be,

$\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}+\mathrm{Ri}=\mathrm{E}\left(\mathrm{t}\right)$… (1)

To obtain the current, since the initial current is $\text{I(0) = 0}$. As the values are$\text{L = 0.1 henry,R = 50 ohms,}$ and $\text{E = 30 volt}$.

$\begin{array}{rcl}0.1\frac{\mathrm{di}}{\mathrm{dt}}+50\mathrm{i}& =& 30\times 10\\ \frac{\mathrm{di}}{\mathrm{dt}}+500\mathrm{i}& =& 300\end{array}$

$\frac{\text{di}}{\text{dt}}\text{= 300 - 500in}$… (2)

Let separate variables and do integration be,

$\begin{array}{rcl}& & \frac{\text{di}}{\text{300 - 500i}}\text{= dt}\\ & & \int \frac{\text{1}}{\text{300 - 500i}}\text{di =}\int \text{dt}\\ & & \text{-}\frac{\text{1}}{\text{500}}\int \frac{\text{- 500}}{\text{300 - 500i}}\text{di =}\int \text{dt}\end{array}$

$$\begin{gathered} -\frac{\mathrm{1}}{\mathrm{500}}\ln (300\; -\; 500\mathrm{i})\; =\mathrm{t\; +}{\mathrm{c}}_{\mathrm{1}} \\ \ln (300\; -\; 500\mathrm{i})\; =\; -\; 500\mathrm{t}+{\mathrm{c}}_{\mathrm{2}} \\ {\mathrm{e}}^{\ln (300\; -\; 500\mathrm{i})}={\mathrm{e}}^{\left(-\; 500\mathrm{t}+{\mathrm{c}}_{\mathrm{2}}\right)} \\ \mathrm{300\; -\; 500}\mathrm{i}={\mathrm{e}}^{{\mathrm{c}}_{\mathrm{2}}}{\mathrm{e}}^{-\; 500\mathrm{t}} \\ \mathrm{300\; -\; 500}\mathrm{i}={\mathrm{ce}}^{-\; 500\mathrm{t}} \\ 500\mathrm{i}=\; 300\; -{\mathrm{ce}}^{-\; 500\mathrm{t}} \end{gathered}$$

$\text{i(t) =}\frac{\text{3}}{\text{5}}{\text{- ke}}^{\text{- 500t}}$… (3)

Obtain the values of constants.

To find the values of constants, apply the point$\text{(i,t) = (0,0)}$in the equation (3), then

$\begin{array}{rcl}& & \text{0amp =}\frac{\text{3}}{\text{5}}{\text{- ke}}^{\text{0}}\\ & & \text{k =}\frac{\text{3}}{\text{5}}\end{array}$

Substitute the value of $k$in the equation (3).

$\text{i(t) =}\frac{\text{3}}{\text{5}}\text{-}\frac{\text{3}}{\text{5}}{\text{e}}^{\text{- 500t}}$… (4)

Obtain the current of LR circuit.

Substitute the valueinto the equation (4).

$\begin{array}{rcl}& & \text{i =}\frac{\text{3}}{\text{5}}{\text{- ke}}^{\text{-}\infty }\\ & & \text{=}\frac{\text{3}}{\text{5}}\text{- 0}\\ & & \text{i =}\frac{\text{3}}{\text{5}}\text{amp}\end{array}$