76.7k views
5 votes
A 30. 0 μF capacitor initially charged to 30. 0 μC is discharged through a 1. 70 kΩ resistor. How long does it take to reduce the capacitor's charge to 30. 0 μC ?

User Jon Glazer
by
7.2k points

1 Answer

2 votes

Answer:

We can use the formula for the discharge of a capacitor through a resistor:

Q(t) = Q0 * e^(-t/(RC))

where Q(t) is the charge on the capacitor at time t, Q0 is the initial charge on the capacitor, R is the resistance, C is the capacitance, and e is the mathematical constant e.

Setting Q(t) to 30.0 μC, Q0 to 30.0 μC, R to 1.70 kΩ, and C to 30.0 μF, we get:

30.0 μC = 30.0 μC * e^(-t/(1.70 kΩ * 30.0 μF))

Simplifying, we get:

1 = e^(-t/(51.0 s))

Taking the natural logarithm of both sides, we get:

ln(1) = ln(e^(-t/(51.0 s)))

0 = -t/(51.0 s)

Solving for t, we get:

t = 0 s

This means that the capacitor is already discharged to 30.0 μC, so it took no time for this to happen.

User David Terei
by
8.3k points