1. Consider the circuit shown, with three ideal batteries and three resistors. U

Question

1. Consider the circuit shown, with three ideal batteries and three resistors. Use Kirchhoff?s Laws to: a. Determine the direction of the current through each resistor (i.e., up or down). b. Determine the power dissipated in each resistor. 2. In the circuit shown, the capacitor has capacitance 20 mF and each resistor has resistance 50 Omega. 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

1)Assuming the current in the left loop is x clockwise and right loop is y anti-clockwise

By Kirchoff's law,

12-10x-20(x+y)-3=0

6-60y-20(x+y)-3=0

Solving simultaneously,

x=0.33A and y=-0.045A

This means that the current in 10 ohm resistor is upward and 0.33A

Current in 20 ohm resistor is upward and (0.33-0.045)=0.285A downward

Current in 60 ohm resistor is downward and 0.045A

b)P10=10*0.33^2=1.089W

P20=20*0.285^2=1.625W

P60=60*0.045^2=0.122W

2)Equivalent resistance of circuit=75 ohm (50+50/2)

Time constant=RC=0.02*50=1s

I=V0/R*e^-t/RC=2.5/(75)*e^-t

I=0.0333e^-t

I(1)=0.0333e^-1=0.01225A

P(A)=0.01225^2*50=0.0075W

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