Question

A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed at t = 0 s. At t = 10.0 s, the potential difference across the capacitor is 1.64 V. (a) What is the time constant of the circuit? (b) What is the potential difference across the capacitor at t = 18.8 s?

Answer 1

Answer:

• a. $  \tau =  2.1161 s$

• b. $V(18.8 \ s) = 0.0256 \ V$

Explanation:

a.

The equation for the voltage V of  discharging capacitor in an RC circuit at time t is:

$V(t) = V_0 e^{(- \frac{t}{\tau}) }$

where $V_0$ is the initial voltage, and $\tau$ is the time constant.

For our problem, we know

$V_0 = 185 \ V$

and

$V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

So

$185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$ e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }$

$ ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })$

$ - \frac{10 \ s}{\tau}  = ln (\frac{1.64 \ V}{ 185 \ V })$

$  \tau =  \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }$

This gives us

$  \tau =  2.1161 s$

and this is the time constant.

b.

At t = 18.8 s we got:

$V(18.8 \ s) = 185 \ V  \ e^{(- \frac{18.8 \ s}{2.1161 s}) }$

$V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }$

$V(18.8 \ s) = 0.0256 \ V$