Question

A capacitor with an initial potential difference of 116 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 1.54 V. (a) What is the time constant of the circuit? (b) What is the potential difference across the capacitor at t = 16.1 s?

Answer 1

Answer:

(a). The time constant of the circuit is 2.10

(b). The potential difference across the capacitor at t=17.0 s is 0.0543 V.

Explanation:

Given that,

Initial potential difference = 116 V

Potential difference across the capacitor = 1.54 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

$1.00=116e^{\dfrac{-10.0}{RC}}$

$0.0086=e^{\dfrac{-10.0}{RC}}$

On taking ln

$ln(0.0086)=\dfrac{-10}{RC}$

$RC=\dfrac{-10}{ln(0.0086)}\\RC=2.10$

(b). We need to calculate the potential difference across the capacitor at t=16.1 s

Using formula again

$V(16.1)=116e^{\dfrac{-16.1}{2.10}}$

$V{16.1}=0.0543\ V$

Hence, (a). The time constant of the circuit is 2.10

(b). The potential difference across the capacitor at t=17.0 s is 0.0543 V.