EET 102
Fall 1997
Prof. Ken Reid

Quiz 2 - Answers

1) Find the voltage Vx in the given circuit

2) Find the current I

Answers:

No current flows through the 10 ohm resistor (Vx is across an open circuit!)
No current flows through the 8.3 ohm resistor (it is shorted out!)

Therefore, the current I flows through the Voltage source, the 5 ohm resistor, and the 15 ohm resistor, and finally through the wire shorting the 8.3 ohm resistor.

I = V/R = 20 V / (5 ohm + 15 ohm) = 20V / 20 ohm = 1 A

Now, use KVL around the top loop (with Vx). Start where the 5 ohm and the 10 ohm meet, go around clockwise:

Voltage across the 10 ohm R + Vx - Voltage across the 5 ohm resistor = 0

V across the 10 ohm = 0 [IT HAS NO CURRENT; V=RI; V=0]
0 + Vx = V across the 5 ohm resistor

V = R*I = 5 ohm * 1 A = 5 V

So, Vx = 5V

3) What is the power dissipated in the 8.3 ohm resistor?

P = VI :Notice: V = 0 AND I = 0, so P = 0
P = (I**2)R :NOTICE: I=0 so P=0
P = (V**2)/R :NOTICE: V=0 so P=0
The 8.3 ohm resistor dissipates 0 W - it is effectively out of the curcuit.