In the circuit shown, the capacitor is initially uncharged, and V = 10 V. At time t = 0, switch S is closed. If $ \tau $denotes the time constant, the approximate current through the 3 $\Omega$ resistor when t = $ \tau $/10 is:
A) 0.50 A
B) 0.75 A
C) 1.0 A
D) 1.5 A
E) 3.0 A

Question

In the circuit shown, the capacitor is initially uncharged, and V = 10 V. At time t = 0, switch S is closed. If  $ \tau $denotes the time constant, the approximate current through the 3  $\Omega$  resistor when t =  $ \tau $/10 is: 
A) 0.50 A 
B) 0.75 A 
C) 1.0 A 
D) 1.5 A 
E) 3.0 A

Quizplus · Accepted Answer

The time constant for the circuit is given as F1F1F1S1F1F1F10 = R*C = 3*2 = 6s.
At t = F1F1F1S1F1F1F10/10 = 0.6s, the capacitor would have charged to approximately 63.2% of its maximum charge.
Therefore, the voltage across the capacitor at this time is Vc = V*(1 - e^(-t/RC)) = 10*(1 - e^(-0.6/6)) = 4.64V.
The current through the 3 F1F1F1S1F1F1F10 resistor is given by I = Vc/R = 4.64/3 = 1.547 A.
Approximating to one significant figure, we get the answer as 1.0 A. Therefore, the correct choice is (C).

In the Circuit Shown, the Capacitor Is Initially Uncharged, and V