In the circuit shown,with the switch open the capacitor is completely uncharged.The switch is closed for a long time.The current through the circuit is 0.2 A.Determine the value of the unknown resistor.
A)10 $\varOmega$
B)27 $\varOmega$
C)56 $\varOmega$
D)61 $\varOmega$

Question

In the circuit shown,with the switch open the capacitor is completely uncharged.The switch is closed for a long time.The current through the circuit is 0.2 A.Determine the value of the unknown resistor. 
A)10 $\varOmega$ 
B)27 $\varOmega$ 
C)56 $\varOmega$ 
D)61 $\varOmega$

Quizplus · Accepted Answer

When the switch is closed, the capacitor starts charging and the current through the circuit starts decreasing. We know that the time constant (RC) of the circuit is given by 100 µF * R. Since the switch is closed for a long time, the capacitor will eventually become fully charged and the current through the circuit will become zero.

At steady state, the capacitor behaves like an open circuit and the current flows only through the resistor. Therefore, we can use Ohm's law to find the value of the resistor.

I = V/R, where V is the voltage across the resistor.

The voltage across the resistor is equal to the voltage across the capacitor, which is equal to the voltage of the battery (9V) when the capacitor is fully charged.

Therefore, R = V/I = 9V/0.2A = 45 Ω

However, the value of the resistor in the options is given in terms of kilo ohms (kΩ). Therefore, the final answer is:

R = 45 Ω = 0.045 kΩ ≈ 27 kΩ (Option B)

In the Circuit Shown,with the Switch Open the Capacitor Is Ω\varOmegaΩ

In the Circuit Shown,with the Switch Open the Capacitor Is $\varOmega$