Question

A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 60 Ω, R3 = 49 Ω and R4 = 133 Ω. The capacitance is C = 75 μF and the battery voltage is V = 24 V.

1)

The switch has been open for a long time when at time t = 0, the switch is closed. What is I₁(0), the magnitude of the current through the resistor R₁ just after the switch is closed?

2) What is I₁(∞), the magnitude of the current that flows through the resistor R₁ a very long time after the switch has been closed?

3)

What is Q(∞), the charge on the capacitor after the switch has been closed for a very long time?

4)

R3 R5 Ra

Consider the circuit above, with R₅ = 123 Ω in series with the capacitor. Once again, the switch has been open for a long time when at time t = 0, the switch is closed. What is I₁(0), the magnitude of the current through the resistor R₁ just after the switch is closed?

5)

Continuing with the new circuit, what is Q(∞), the charge on the capacitor after the switch has been closed for a very long time?

Answer 1

When time t = 0, the switch is closed, capacitor acts as short circuit so By Applying kirchott rate, v - I_1 R_1 = I_1 R_4 = 0 I_1 = v/R_1 + R_4 = 24 v/60 Ohm + 135 Ohm = 0.12435 A After a very long time, the capacitor acts as open circuit, so, v - I_1 R_1 - I_1 R_2 - I_1 R_3 - I_1 R_4 = 0 I_1 = v/R_1 + R_2 + R_3 + R_4 = 24 v/60 Ohm + 60 Ohm + 49 Ohm + 133 Ohm I_1 = 0.07947 A at t rightarrow infinity, I_1 = I_4 By Applying loop equation v - I_1 R_1 - v_0 I_4 R_4 = 0 v_0 = 24 v - (60 Ohm + 133 Ohm) (0.07947 A) v_0 = 8.66229 v - 8.66 v Q = v_0 c = (8.66229 v) (75 mu F) Q = 649.67 mu c At t = 0, R_23 = (60 + 49) Ohm = 109