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 $$${I}_{{s}}$$$).