Question

A 591 μF capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of 88.5 V to 11.9 V in 3.09 s. Find the resistance of the resistor in kilohms.

Answer 1

The resistance of the resistor is 2.605 kilo ohms.

What is resistance?

The resistance is defined as a measure of the opposition to current flow in an electrical circuit.

Given that the capacitance of the capacitor is 591 μF. The initial potential difference Vi is 88.5 V and the final potential difference Vf is 11.9 V. The time is 3.09 s.

Let's consider that R is the resistor, then

$V_f = V_i e^{(\dfrac {-t}{RC})}$

$e^ {(\dfrac {-t}{RC})} = \dfrac {V_f}{V_i}$

$e^ {(\dfrac {-t}{RC})} = \dfrac {11.9}{88.5}$

$e^ {(\dfrac {-t}{RC})} = 0.135$

Taking natural log on both sides, we get,

$\dfrac {-t}{RC} = In \;0.135$

$RC = \dfrac {-3.09}{-2.002}$

$RC = 1.54$

$R = \dfrac {1.54}{591 \times 10^{-6}}$

$R = 2605.75$

Hence we can conclude that the resistance of the resistor is 2.605 kilo ohms.

To know more about the resistance, follow the link given below.

https://brainly.com/question/4289257.