Question

Referring to Figure 21.38, draw a graph of potential difference across the resistor versus time, showing at least two intervals of\(\tau \). Also draw a graph of current versus time for this situation.

Short answer

The potential difference across the resistor versus time, \({{\rm{V}}_{\rm{R}}}{\rm{(t)}} = - {\rm{V}}{{\rm{e}}^{{\rm{ - t/RC}}}}\)

Current versus time graph, \({\rm{I(t)}} = \frac{{\rm{V}}}{{\rm{R}}}{{\rm{e}}^{{\rm{ - t/RC}}}}\)

Step-by-step solution

Voltage across the capacitor

In an RC circuit with a capacitor initially uncharged, the voltage across the capacitor as a function of time is given by:

\({\rm{V = emf}}\left( {{\rm{1 - }}{{\rm{e}}^{\frac{{{\rm{ - 1}}}}{{{\rm{RC}}}}}}} \right)\)

Here,\({\rm{V}}\)is the voltage across the terminals of the capacitor due to charging, emf is the voltage across the terminals of the battery used in the circuit,\({\rm{t}}\)is time calculated from the start of the charging process,\({\rm{R}}\)is the resistance in the circuit, and\({\rm{C}}\)is the value for the capacitance of the capacitor.

The current can be obtained using the relation\({\rm{I = V/R}}\).

Potential difference across the resistance

Consider an RC circuit and calculate the time dependence of voltage difference across the resistor.

The voltage on the capacitor is

\({{\rm{V}}_{\rm{C}}} = {\rm{V}}\left( {{\rm{1}} - {{\rm{e}}^{{\rm{ - t/RC}}}}} \right)\)

Use the second Kirchhoff rule (loop rule),

\(\begin{align}{{}{}}{{\rm{V + }}{{\rm{V}}_{\rm{R}}}{\rm{ - }}{{\rm{V}}_{\rm{C}}} = {\rm{0}}}\\{{{\rm{V}}_{\rm{R}}} = {{\rm{V}}_{\rm{C}}} - {\rm{V}}}\\{{{\rm{V}}_{\rm{R}}}{\rm{(t)}} = - {\rm{E}}{{\rm{e}}^{{\rm{ - t/RC}}}}}\end{align}\)

where \({\rm{E}}\) is the emf of the battery. As expected, the potential difference across the resistor in the current direction is negative.

The magnitude of the voltage difference is plotted below for \({\rm{E}} = {\rm{10}}\;{\rm{V}}\)and \({\rm{R}} = {\rm{2}}\;\Omega \).