## PHYS345 Electricity and Electronics

Quiz 2 Solution: dc Circuit Analysis

What is the voltage across the current source?

Via mesh analysis:

Defining the mesh currents in the conventional way, the KVL equation for mesh 1 is:
5 I1 - 3 I2 = 2.

KVL for mesh 2:
-3 I1 +9 I2 - 5 I3 = 0

By inspection of mesh 3:
I3 = -2

Subsequent elimination of I3 in the mesh 2 equation leads to:
-3 I1 +9 I2 = -10

Multiplying the equation for mesh 1 by 3 yields:
15 I1 - 9 I2 = 6.

Adding the two equations immediately above produces:
I1 = -1/3 A = -0.33 A.

Back substitution of I1 into one of the equations involving both I1 and I2 produces:
I2 = -11/9 A = -1.22 A.

A straightforward way to find the voltage across the current source is to evaluate the voltage across the 5 ohm resistor. The current through the resistor is I2 - I3 = 0.78 A, thus the voltage across the resistor must be (0.78 A)(5 ohm) = 3.89 V.

"http://www.physics.udel.edu/~watson/phys345/quiz/02soln1.html"
Last updated Sept. 21, 1999.
Copyright George Watson, Univ. of Delaware, 1999.