Consider the circuit shown, with three ideal batteries and three resistors. Use
ID: 1346358 • Letter: C
Question
Consider the circuit shown, with three ideal batteries and three resistors. Use Kirchhoff's Laws to: Determine the direction of the current through each resistor (i.e., up or down). Determine the power dissipated in each resistor. In the circuit shown, the capacitor has capacitance 20 mF and each resistor has resistance 50 ohm. The capacitor is charged to 2.5 V and at time t = 0 s, it is connected to the circuit and the capacitor begins to discharge as shown. What is the instantaneous power dissipated in Resistor A at time t = 1.0 s?Explanation / Answer
2. Assume directions of current as shown in figure,
a )Applying KCl at node point a
I1= I2+I3
I3= I1-I2---------------(1)
Applying KVL to loop1 ,
12-10I1-20I2-3=0 -------------(2)
Applying KVL to loop2 ,
12-10I1-60I3-6=0 -------------(3)
From (1)
12-10I1-60(I1-I2)-6=0
12-10I1-60I1+60I2-6=0
12-70I1+60I2-6=0-----------------(4)
Simplifying simultaneous equations (2) and (4) we get,
I1=0.33A and I2=0.285A
I3= I1-I2= 0.33-0.285 = 0.045 A
Directions of I1 , I2and I3 are same as shown in the diagram
b) P1= I1R1 = 0.33*10 = 3.3J
P2=I2*R2= 0.285*20 = 5.7 J
P3=I3*R3= 0.045*60 = 2.7J
3. C= 20mF ,
Requ = 50 ohm + ( 50ohm || 50 ohm) = 50ohm +25ohm = 75ohm
Current at t= 1s
It=1= (V/Requ)*et/RC = (2.5/75)e1/(75*20*10^-3) = 0.065 A
Power in resistor A = I2R = 0.065*50 = 3.25 J