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Chapter 4: Q26-80P (page 879)

You set up the circuit shown in Fig. 26.22a, where \(R = 196\;\Omega \). You close the switch at time \(t = 0\) and measure the magnitude \(i\) of the current in the resistor \(R\)as a function of time \(t\) since the switch was closed. Your results are shown in Fig. P26.80, where you have chosen to plot \({\rm{ln}}\;i\)as a function of \(t\).

(a) Explain why your data points lie close to a straight line.

(b) Use the graph in Fig. P26.80 to calculate the capacitance \(C\) and the initial charge \({Q_0}\) on the capacitor.

(c) When \(i = 0.05\;{\rm{A}}\), what is the charge on the capacitor?

(d) When \(q = 0.5 \times {10^{ - 4}}\;{\rm{C}}\), what is the current in the resistor?

Short Answer

Expert verified

(a) The graph of log of current verses time is a straight line with slope \( - \frac{1}{{RC}}\). Thus, the data points lie close to a straight line.

Step by step solution

01

Given data

Resistance in the circuit is

.\(R = 196\;\Omega \).

From the graph, log of current at time \({t_i} = 1.5\;{\rm{ms}}\)

\({\rm{ln }}{i_i} = - 3\).

From the graph, log of current at time \({t_f} = 3\;{\rm{ms}}\)

\({\rm{ln }}{i_f} = - 4\)

02

Current in a CR circuit

Current in the circuit after time \(t\) when a capacitor of initial charge \({Q_0}\) and capacitance \(C\) is connected to a resistance \(R\) is:

\(i = \frac{{{Q_0}}}{{RC}}{e^{ - \frac{t}{{RC}}}}\) .....(i)

03

(a) Determination of nature of the graph between log of current and time

Take log of equation (i) to get

\(\begin{aligned}{\rm{ln }}i = {\rm{ln}}\left( {\frac{{{Q_0}}}{{RC}}{e^{ - \frac{t}{{RC}}}}} \right)\\ = {\rm{ln}}\left( {\frac{{{Q_0}}}{{RC}}} \right) - \frac{t}{{RC}}\end{aligned}\)

Thus, graph of log of current verses time should be a straight line with slope \( - \frac{1}{{RC}}\) .

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