A 1 000-V battery,a 3 000-Ω resistor,and a 0.50-μF capacitor are connected in series with a switch.The time constant for such a circuit,designated by the Greek letter,τ,is defined as the time that the capacitor takes to charge to 63% of its capacity after the switch is closed.What is the current in the circuit at a time interval of τ seconds after the switch has been closed?
A)0.14 A
B)0.21 A
C)0.12 A
D)0.32 A

Question

A 1 000-V battery,a 3 000-Ω resistor,and a 0.50-μF capacitor are connected in series with a switch.The time constant for such a circuit,designated by the Greek letter,τ,is defined as the time that the capacitor takes to charge to 63% of its capacity after the switch is closed.What is the current in the circuit at a time interval of τ seconds after the switch has been closed? 
A)0.14 A 
B)0.21 A 
C)0.12 A 
D)0.32 A

Quizplus · Accepted Answer

The time constant for the circuit is given by:

τ = RC

where R is the resistance and C is the capacitance. Substituting the given values, we get:

τ = (3,000 Ω)(0.50 μF) = 1.5 ms

After a time interval of τ seconds, the capacitor would have charged to 63% of its capacity, which means that the voltage across the capacitor would be 0.63 times the battery voltage. Therefore, the voltage across the resistor would be:

Vr = Vb - Vc = Vb - 0.63Vb = 0.37Vb

Using Ohm's law, we can find the current in the circuit:

I = Vr/R = (0.37Vb)/(3,000 Ω) = (0.37)(1,000 V)/(3,000 Ω) = 0.12 A

Therefore, the answer is (C), 0.12 A.

A 1 000-V Battery,a 3 000-Ω Resistor,and a 0