A circuit you’re using discharges a $20\mu F$capacitor through an unknown resistor. After charging the capacitor, you close a switch at $t=0s$and then monitor the resistor current with an ammeter. Your data are as follows:

Use an appropriate graph of the data to determine

(a) the resis-tance and

(b) the initial capacitor voltage.

We need to use an appropriate graph of the data to determine the resistance.

To discharge the capacitor, the charges flow through the circuit in time $t.$The initial current immediately after the switch closes is given by equation $(28.33)$in form

localid="1649265224749" $I=I_0{e}^{-t/RC}$

This equation rules the graph. The graph is showing below, and if we take the fit for the graph, we find the initial current which is the maximum current at localid="1649096275882" $t=0$to belocalid="1649096282158" $I_0=1340\mu A$and the termlocalid="1649096289868" $(1/RC)=0.77.$

Now, we use the value of C to get the resistance by

localid="1650101276905" $R=\frac{1}{0.77C}=\frac{1}{0.77(20\times {10}^{-6}F)}=65\times {10}^3\Omega$.

We need to determine the initial capacitor voltage.

The initial potential across the capacitor is calculated by,

$∆V_C=\frac{Q_0}{C}=\frac{I_{0}RC}{C}=I_{0}R$

So, let us plug the values for $I_0$and R to get $∆V$,

$∆V_C=I_{0}R=(1340\times {10}^{-6}A)(65\times {10}^3\Omega )=87.0V.$