## PHYS345 Electricity and Electronics

**Hints for 9/9 Assignment**

**Hint for S3.1b**
By inspection, left mesh current *I*_{1} is 2 A.

The right mesh equation is 6 *I*_{x} - 4 *I*_{1} = 10.

**Hint for S3.2b**

By inspection, left mesh current *I*_{1} is 6 A.

By inspection, right mesh current *I*_{3} is - 3 A.

The center mesh equation is 9 *I*_{2} - 3 *I*_{1} - 6 *I*_{3} = 18.

Solve for *I*_{2} and find the voltage difference across either of the center
branches.

**Hint for 3.13**

- Label the mesh currents as follows:
- I
_{1} for the left mesh.
- I
_{2} for the top right mesh.
- I
_{3} for the bottom right mesh.

- The mesh equation for the left loop is then:

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

"http://www.physics.udel.edu/~watson/phys345/protected/exercises/hints/0909.html"

Last updated Sept. 9, 1999.

Copyright George Watson, Univ. of Delaware, 1999.