Question

A capacitor with an initial potential difference of 100 V isdischarged through a resistor when a switch between them is closed at t ! 0.At t ! 10.0s,the potential difference across the capacitor is 1.00V. (a) What is the time constant of the circuit? (b)What is the potential difference across the capacitor at t ! 17.0s?

Answer 1

Answer:

(a). The time constant of the circuit is 2.17.

(b). The potential difference across the capacitor at t=17.0 s is 0.0396 V.

Explanation:

Given that,

Initial potential difference = 100 V

Potential difference across the capacitor = 1.00 V

(a). We need to calculate the time constant of the circuit

Using formula of potential difference

$V(t)=V_{0}e^{\dfrac{-t}{RC}}$

Put the value into the formula

$1.00=100e^{\dfrac{-10.0}{RC}}$

$0.01=e^{\dfrac{-10.0}{RC}}$

On taking ln

$ln(0.01)=\dfrac{-10}{RC}$

$RC=\dfrac{-10}{ln(0.01)}$

$RC=2.17$

(b). We need to calculate the potential difference across the capacitor at t=17.0 s

Using formula again

$V(17)=100e^{\dfrac{-17}{2.17}}$

$V{17}=0.0396\ V$

Hence, (a). The time constant of the circuit is 2.17.

(b). The potential difference across the capacitor at t=17.0 s is 0.0396 V.