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

• a.
  `\tau =  2.1161 s`

• b.
  `V(18.8 \ s) = 0.0256 \ V`

Step-by-step explanation:

a.

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

`V(t) = V_0 e^{(- (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^{(- (10 \ s)/(\tau)) } = 1.64 \ V`

So

`185 \ V \ e^{(- (10 \ s)/(\tau)) } = 1.64 \ V`

`e^{(- (10 \ s)/(\tau)) } = (1.64 \ V)/( 185 \ V )`

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

`- (10 \ s)/(\tau)  = ln ((1.64 \ V)/( 185 \ V ))`

`\tau =  (-10 \s)/(ln ((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^{(- (18.8 \ s)/(2.1161 s)) }`

`V(18.8 \ s) = 185 \ V \ e^{(- (18.8 \ s)/(2.1161 s)) }`

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