The circuit shown in Figure 5 contains two emfs of opposite polarity E1 = 40 V,

Question

The circuit shown in Figure 5 contains two emfs of opposite polarity E1 = 40 V, and E2 = -20 V. The circuit also contains a switch, resistors R1 and R2, and a capacitor C. The capacitor is initially discharged. The switch is connected to position 'a' for a period of 8 seconds, after which it is returned to the unconnected position 'b' for a period of 12 seconds before being connected to position 'c' for a further 8 seconds. W are the time constants for each of the three time intervals? Calculate the potential difference across the capacitor, Vc, at the end of each time interval. Sketch a graph of Vc against time for the whole 28 second period.

Explanation / Answer

a) the time constant is same for all the three intervals and is equal to RC = 680x10^3 x 8x10^(-6) = 5.44secs for positions 'a' and 'c' for position 'b', RC = (680+820)x10^3 x8x10^(-6) = 12secs b) when the switch is in position 'a': this resembles a RC circuit, the voltage across 820kohms is same as 40V apply KVL in the right loop: 40 = 680x10^3i + (1/(8x10^-6))int(i)dt differentiate and solve di/dt + 0.18i = 0 solution to this is i(t) = ce^(-0.18t) c= V/R =40/(680x10^3) = 5.88x10^-5 i(t) = 5.88x10^(-5)e^(-0.18t) the voltage across the capacitor is Vc = (1/C)int(i)dt Vc = (1/(8x10^(-6)))(5.88x10^(-5)e^(-0.18t)/-0.18) + C1 at t=0,C1=V(0) = 40 Vc(8) = 30.32V when switch in position 'b': apply KVL in the right loop: 0 = (680+820)x10^3i + (1/(8x10^-6))int(i)dt differentiate and solve di/dt + 0.083i = 0 solution to this is i(t) = ce^(-0.083t) c= V/R =40/((680+820)x10^3) = 2.67x10^-5 i(t) = 2.67x10^(-5)e^(-0.18t) the voltage across the capacitor is Vc = (1/C)int(i)dt Vc = (1/(8x10^(-6)))(2.67x10^(-5)e^(-0.083t)/-0.083) + C2 at t=0,C2= 30.32 Vc(12) = 15.47V when switch in position 'c': apply KVL in the right loop: 20 = 680x10^3i + (1/(8x10^-6))int(i)dt differentiate and solve di/dt + 0.18i = 0 solution to this is i(t) = ce^(-0.18t) c= V/R =20/(680x10^3) = 2.94x10^-5 i(t) = 2.94x10^(-5)e^(-0.18t) the voltage across the capacitor is Vc = (1/C)int(i)dt Vc = (1/(8x10^(-6)))(2.94x10^(-5)e^(-0.18t)/-0.18) + C3 at t=0,C3=15.47 Vc(8) = 10.63 c) the graph exponentially decreases from 30.32 to 10.63 for 28sec..

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