The following circuit is at steady state for t<0. At t=0 sec, the switch is open. Let R1=10 ohms, R2=14 ohms, R3=4 ohms, C1=2 F, Vx=12 V and Ix=1 A. Find Vc1(1 sec) and voltage across resistor R3 = V(0.7 sec) , as follows:
To find the voltage across capacitor C1 at 1 second (t=1 sec), we first need to find the initial voltage across C1 (Vc1(0)) before the switch is closed.
Using Kirchhoff's Laws, we can find Vc1(0):
Vx = Ix * R1 + Vc1(0)
12 = 1 * 10 + Vc1(0)
Vc1(0) = 2 V
Now, we can use the following equation to find Vc1(1 sec):
Vc1(t) = Vx - Ix * R1 * e^(-t/(R1 * C1))
Vc1(1 sec) = 12 - 1 * 10 * e^(-1/(10 * 2))
Vc1(1 sec) = 12 - 10 * e^(-0.05)
Vc1(1 sec) ≈ 12 - 10 * 0.951
Vc1(1 sec) ≈ 12 - 9.51
Vc1(1 sec) ≈ 2.41 V
Now, to find the voltage across resistor R3 at 0.7 seconds (t=0.7 sec), we use the formula:
V(t) = V_initial * e^(-t/(R * C))
Using the values given:
V(0.7 sec) = V_initial * e^(-0.7/(R3 * C1))
V(0.7 sec) = 12 * e^(-0.7/(4 * 2))
V(0.7 sec) = 12 * e^(-0.175)
V(0.7 sec) ≈ 12 * 0.839
V(0.7 sec) ≈ 10.07 V
Therefore, Vc1(1 sec) ≈ 2.41 V and V(0.7 sec) ≈ 10.07 V.