Question:

A circuit that has a resistor, \(R\), and an inductor, \(L\), which are connected in series with a DC voltage source \(V\). The initial value of the current is \(I(t=0)=0\) A. Find the current as a function of time.

Answer:

From Kirchoff’s law, the generated voltage is consumed by each element.

\[  V-RI-L\frac{dI}{dt}=0  \]

\[  RI+L\frac{dI}{dt} = V  \]

\[  I’ + \frac{1}{\tau}I = \frac{V}{L}  \]

where \(I’=\frac{dI}{dt}\) and \(\tau=\frac{L}{R}\). Now, let \(D\equiv \frac{d}{dt}\).

\[  \left(D+\frac{1}{\tau}\right)I=\frac{V}{L}  \]

The fundamental solution is

\[ I_f = C_1e^{-\frac{t}{\tau}} \]

The particular solution is

\[  I_p = \frac{1}{D+\frac{1}{\tau}}\cdot \frac{V}{L}=\frac{1}{0+\frac{1}{\tau}}\cdot \frac{V}{L}=\frac{V}{R}  \]

Therefore, we have the general solution.

\[  I = C_1e^{-\frac{t}{\tau}} + \frac{V}{R}  \]

When \(t=0\), \(I=0\). So \(0=C_1+V/R\) and \(C_1=-V/R\). Then, the current is expressed as

\[  I(t) = -\frac{V}{R}e^{-\frac{t}{\tau}}+\frac{V}{R}  \]

or

\[  I(t) = \frac{V}{R}(1-e^{-\frac{t}{\tau}})  \]