Question

Dielectric materials used in the manufacture of capacitors are characterized by conductivities that are small but not zero. Therefore, a charged capacitor slowly loses its charge by "leaking" across the dielectric. If a certain 3.25 µF capacitor leaks charge such that the potential difference decreases to half its initial value in 4 s, what is the equivalent resistance of the dielectric?

Answer 1

Answer:

Equivalent resistance of the dielectric= R =1.78 × 10^6 Ω

Explanation:

This is just an RC circuit, where

q = CVoe-t/RC.

V is proportional to q; so if V has lost half its value, so has q.

So we need R such that

e-t/RC = 0.5

Taking log on both sides, we get

-t/RC = ln(0.5)

R = -t/ln(0.5)C

= (-4 s)/[(3.25×10^-6 F)(-0.693147)]

= 1.78 × 10^6 Ω