Question

A 50 mF capacitor that had been charged to 30 V is dis-charged through a resistor. FIG­URE P28.70 shows the capacitor voltage as a function of time. What is the value of the resistance?

Short answer

$R=72.8\Omega$

Step-by-step solution

Given information.

We need to find the values for R.

Simplification.

When the switch is off, the capacitor discharge and the current flows through the resistor. To discharge the capacitor, the charges flow through the circuit in time $t$.the initial voltage is related to the final voltage by

$∆V=∆V_0{e}^{-t/RC}$

where R is the resistance and $C$is the capacitance. Let us solve equation (1) for localid="1649245722256" $R$to get it by

$R=\frac{-t}{CIn(∆V/Q∆V_0)}$

At the time localid="1649245730682" $t$=$4ms$,the final voltage islocalid="1649245738870" $∆V=10v$. We plug the values for localid="1649245747283" $t$,$C,∆V$and $∆V_0$into equation to get R

$$\begin{gathered} R=\frac{-t}{CIn(∆V/Q∆V_0)} \\ =\frac{-4\times {10}^{-3}s}{(50\times {10}^{-6}F)\left(In\right(10/30\left)\right)} \\ =72.8\Omega \end{gathered}$$.