Electrical Circuits

The basic equation governing the amount of current ${I}$ (in amperes) in a simple RL circuit consisting of a resistance ${R}$ (in ohms), an inductance ${L}$ (in henries), and an electromotive force (abbreviated as 'emf') ${E}$ (in volts) is $\frac{{{d}{I}}}{{{d}{t}}}+\frac{{R}}{{L}}{I}=\frac{{E}}{{L}}$ .
For an RC circuit consisting of a resistance, a capacitance ${C}$ (in farads), an emf, and no inductance, the equation governing the amount of electrical charge ${q}$ (in coulombs) on the capacitor is $\frac{{{d}{q}}}{{{d}{t}}}+\frac{{1}}{{{R}{C}}}{q}=\frac{{E}}{{R}}$ . The relationship between ${q}$ and ${I}$ is ${I}=\frac{{{d}{q}}}{{{d}{t}}}$ .

Example 1. An RL circuit has an emf of 5 volts, a resistance of 50 ohms, an inductance of 1 henry, and no initial current. Find the current in the circuit at any time ${t}$ .

Here, ${E}={5}$ , ${R}={50}$ , and ${L}={1}$ ; hence, the differential equation becomes $\frac{{{d}{I}}}{{{d}{t}}}+{50}{I}={5}$ . This is a linear equation; its solution is ${I}={c}{{e}}^{{-{50}{t}}}+\frac{{1}}{{10}}$ .

At ${t}={0}$ , ${I}={0}$ ; thus, ${0}={c}{{e}}^{{-{50}\cdot{0}}}+\frac{{1}}{{10}}$ , or ${c}=-\frac{{1}}{{10}}$ . The current at any time, then, is ${I}=-\frac{{1}}{{10}}{{e}}^{{-{50}{t}}}+\frac{{1}}{{10}}$ .

The quantity $-\frac{{1}}{{10}}{{e}}^{{-{50}{t}}}$ is called transient current, since this quantity goes to zero ("dies out") as ${t}\to\infty$ .

The quantity $\frac{{1}}{{10}}$ is called steady-state current. As ${t}\to\infty$ , the current ${I}$ approaches the value of steady-state current.

Let's work another helpful example.

Example 2. An RC circuit has an emf of ${300}{\cos{{\left({2}{t}\right)}}}$ volts, a resistance of 150 ohms, a capacitance of $\frac{{1}}{{600}}$ farad, and an initial charge on the capacitor of 5 coulombs. Find the charge on the capacitor at any time ${t}$ and the steady-state current.

Here, ${E}={300}{\cos{{\left({2}{t}\right)}}}$ , ${R}={150}$ , and ${C}=\frac{{1}}{{600}}$ .

So, $\frac{{{d}{q}}}{{{d}{t}}}+{4}{q}={2}{\cos{{\left({2}{t}\right)}}}$ . This is a linear differential equation. It can be rewritten as $\frac{{{d}{q}}}{{{d}{t}}}{{e}}^{{{4}{t}}}+{4}{{e}}^{{{4}{t}}}{q}={2}{{e}}^{{{4}{t}}}{\cos{{\left({2}{t}\right)}}}$ , or $\frac{{{d}{\left({q}{{e}}^{{{4}{t}}}\right)}}}{{{d}{t}}}={2}{{e}}^{{{4}{t}}}{\cos{{\left({2}{t}\right)}}}$ .

Integrating both sides gives ${q}{{e}}^{{{4}{t}}}=\int{\left({2}{{e}}^{{{4}{t}}}{\cos{{\left({2}{t}\right)}}}\right)}{d}{t}$ . Using integration by parts, we obtain that ${q}{{e}}^{{{4}{t}}}={c}+\frac{{{{e}}^{{{4}{t}}}}}{{5}}{\left({\sin{{\left({2}{t}\right)}}}+{2}{\cos{{\left({2}{t}\right)}}}\right)}$ .

The charge on the capacitor at any time ${t}$ is ${q}={c}{{e}}^{{-{4}{t}}}+\frac{{1}}{{5}}{\left({\sin{{\left({2}{t}\right)}}}+{2}{\cos{{\left({2}{t}\right)}}}\right)}$ .

Since ${q}={5}$ when ${t}={0}$ , we have that ${5}={c}{{e}}^{{-{4}\cdot{0}}}+\frac{{1}}{{5}}{\left({\sin{{\left({2}\cdot{0}\right)}}}+{2}\cdot{\cos{{\left({2}\cdot{0}\right)}}}\right)}$ , or ${c}=\frac{{23}}{{5}}$ .

Thus, ${q}=\frac{{1}}{{5}}{\left({23}{{e}}^{{-{4}{t}}}+{\sin{{\left({2}{t}\right)}}}+{2}{\cos{{\left({2}{t}\right)}}}\right)}$ .

Next, since ${I}=\frac{{{d}{q}}}{{{d}{t}}}$ , we have that ${I}=\frac{{1}}{{5}}{\left(-{92}{{e}}^{{-{4}{t}}}+{2}{\cos{{\left({2}{t}\right)}}}-{4}{\sin{{\left({2}{t}\right)}}}\right)}$ , and the steady-state current is ${I}_{{s}}=\frac{{1}}{{5}}{\left({2}{\cos{{\left({2}{t}\right)}}}-{4}{\sin{{\left({2}{t}\right)}}}\right)}$ ( $-\frac{{92}}{{5}}{{e}}^{{-{4}{t}}}$ is transient because this quantity goes to zero as ${t}\to\infty$ , unlike ${I}_{{s}}$ ).