## 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 I
_{1} - 3 I_{2} = 2.

- KVL for mesh 2:

- -3 I
_{1} +9 I_{2} - 5 I_{3} = 0

- By inspection of mesh 3:

- I
_{3} = -2

- Subsequent elimination of I
_{3} in the mesh 2 equation leads to:

- -3 I
_{1} +9 I_{2} = -10

- Multiplying the equation for mesh 1 by 3 yields:

- 15 I
_{1} - 9 I_{2} = 6.

- Adding the two equations immediately above produces:

- I
_{1} = -1/3 A = -0.33 A.

- Back substitution of I
_{1} into one of the equations involving both I_{1} and I_{2} produces:
- I
_{2} = -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 I_{2} - I_{3} = 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.