Consider an RC circuit with a capacitance of 150 $\mu$0F,a resistance of 1 MΩ and connected to a 200-V power source for a long time.The power supply is disconnected and shorted.How long should one wait before the voltage across the capacitor drops to below 50 V?
A)52 s
B)104 s
C)176 s
D)208 s
E)314 s

Question

Quizplus · Accepted Answer

The time constant of an RC circuit is given by τ = RC, where R is the resistance in ohms (Ω) and C is the capacitance in farads (F). For this circuit, τ = (1 MΩ)(150 μF) = (1×10^6 Ω)(150×10^-6 F) = 150 seconds. The voltage across a discharging capacitor at any time t can be found using the formula V(t) = V0e^(-t/τ), where V0 is the initial voltage and V(t) is the voltage at time t. To find the time when the voltage drops to below 50 V from an initial 200 V, we set V(t) = 50 and solve for t: 50 = 200e^(-t/150). Dividing both sides by 200 gives 0.25 = e^(-t/150). Taking the natural logarithm of both sides gives ln(0.25) = -t/150, which simplifies to t = -150 ln(0.25). This calculation yields t ≈ 207.94 seconds, which rounds to approximately 208 seconds.

Consider an RC Circuit with a Capacitance of 150 μ\muμ 0F,a Resistance of 1 MΩ and Connected to a 200-V

Consider an RC Circuit with a Capacitance of 150 $\mu$ 0F,a Resistance of 1 MΩ and Connected to a 200-V