Question

An RC circuit has a resistor R=500Ω, a capacitor of C=2μF, and a battery whose emf is ε= 9 V. The Capacitor is charging. a.) What is the time constant of the circuit? b.) At what time, t, is the charge at 25% of its final value? c.) What is the voltage after a time of 1.5 ms ? d.) What is the current at the same time in part b?

Answer 1

In an RC circuit with R=500Ω, C=2μF, and ε=9V: a) The time constant is 1ms. b) At t = T/4, the charge is 25% of its final value. c) At t = 1.5ms, the voltage is about 8.482V. d) At t = T/4, the current is about 0.368(ε/R).

Step-by-step explanation:

a.) The time constant (T) of an RC circuit is given by the equation T = RC, where R is the resistance in ohms and C is the capacitance in farads. In this case, R = 500Ω and C = 2μF. So, the time constant is T = 500Ω * 2μF = 1ms.

b.) The charge on a capacitor at time t in an RC circuit can be calculated using the equation Q(t) = Q0 * (1 - e^(-t/T)), where Q0 is the initial charge and e is the base of natural logarithm. At t = T/4, the charge is 25% of its final value. So, plugging in the values, we get Q(T/4) = Q0 * (1 - e^(-T/4T)) = Q0 * (1 - e^(-1/4)) = 0.579Q0.

c.) The voltage across the capacitor in an RC circuit can be calculated using the equation V(t) = ε * (1 - e^(-t/T)), where ε is the emf of the battery. At t = 1.5ms, plugging in the values, we get V(1.5ms) = 9V * (1 - e^(-1.5ms/1ms)) ≈ 8.482V.

d.) The current in an RC circuit at time t can be calculated using the equation I(t) = ε/R * e^(-t/T). At t = T/4, plugging in the values, we get I(T/4) = ε/R * e^(-T/4T) = ε/R * e^(-1/4) ≈ 0.368(ε/R).