A series circuit consists of a 12.0 V source of emf, a 2.00 mF capacitor, a 1000 Ω resistor, and a switch. When the switch is closed, how long does it take for the current to reach one-half its maximum value?
A) 0.693 s
B) 1.39 s
C) 2.00 s
D) 1.69 s
E) 0.0250 s

Question

Quizplus · Accepted Answer

We can use the time constant formula for a series RC circuit:

τ = RC

where R is the resistance and C is the capacitance.

The maximum current will be given by:

I_max = EMF/R

Since the circuit is a series circuit, the current through the resistor and the capacitor is the same at any given time.

When the switch is closed, the capacitor will initially act like a short circuit and the current will be maximum. As the capacitor charges up, the current will decrease until it reaches its final steady-state value of I_max.

When the current is half its maximum value, we can relate it to the time constant τ and the time t as follows:

I = I_max * exp(-t/τ)
0.5 I_max = I_max * exp(-t/τ)
0.5 = exp(-t/τ)
ln(0.5) = -t/τ
t = -τ * ln(0.5)

Substituting the given values gives:

τ = RC = (1000 Ω) x (2.00 mF) = 2 s
t = -2 s x ln(0.5) = 1.39 s

Therefore, the answer is B.

A Series Circuit Consists of a 12