During a physics demonstration,a fully charged 98-µF capacitor is discharged through a 125-$\varOmega$ resistor.How long will it take for the capacitor to lose 75% of its initial energy?
A)4.3 ms
B)6.2 ms
C)8.5 ms
D)11 ms
E)24 ms

Question

Quizplus · Accepted Answer

The time constant of the RC circuit is given by the product of the resistance and the capacitance:

τ = RC = (125 Ω)(98 μF) = 12.25 ms

To find the time it takes for the capacitor to lose 75% of its initial energy, we use the formula:

E(t) = E0 e^(-t/τ)

where E0 is the initial energy and E(t) is the energy at time t.

Taking the natural logarithm of both sides and solving for t, we get:

ln(E(t)/E0) = -t/τ

t = -τ ln(E(t)/E0)

We want to find the time it takes for the energy to decrease to 25% of its initial value, so E(t)/E0 = 0.25. Substituting into the equation above, we get:

t = -τ ln(0.25) = -12.25 ms ln(0.25) ≈ 8.5 ms

Therefore, the answer is choice C, 8.5 ms.

During a Physics Demonstration,a Fully Charged 98-µF Capacitor Is Discharged Ω\varOmegaΩ

During a Physics Demonstration,a Fully Charged 98-µF Capacitor Is Discharged $\varOmega$