**Unit 1**

1270 **PRACTICE
EXAM SOLUTION Prob 2**

**2.** (30
points)

Derive an expression for i3. The expression must not contain more than the circuit parameters Va, Vb, ia, R1, R2, and R3.

**ans: **

**sol'n: **Using
passive sign convention, label voltage drop and current measurement polarities.

Use Kirchhoff's laws:

sum v drops around loop = 0

sum i out of node = 0

v drops for loop on left, using Ohm's law for v1 and v2:

Middle loop would include current source, so use slightly larger loop with R2 on left and Vb on right:

Now sum currents out of top node (that consists of the two top nodes connected by a wire). Note: We are always allowed to combine nodes connected by wires.

-i1 + i2 + ia + i3 = 0 A

We now have three equations in three unknowns. We solve for i3. Use the second equation to eliminate i2:

Use the first equation to eliminate i1:

Substitute for i1 and i2 in the third equation:

Solve for i3:

Multiply both sides by R1R2 to clear fractions:

or

**Now
for consistency checks to verify our answer.**
(Optional)

1)** **Consider ia = 0, Vb = 0, and R3
= 0:

Since R2 is bypassed by a short, no current flows in R2. Therefore, we can remove R2 without changing i3:

Our formula gives i3 = .

2)** **Consider ia = 0 (open circuit) and R2 = ´
(open circuit):

Removing R2 and ia leaves total voltage Va + Vb across R1 + R3 in outside loop.

Therefore, we have

For our formula, we use the following identities:

Making these substitutions in our formula gives

(3)** **Consider Va = 0, ia = 0:

Our formula gives or .

(4)** **Consider Va = 0, Vb = 0:

By current divider formula, we have .

** **

Our formula gives or

** **

(5)** **Consider ia = 0, Vb = 0.

Now

Our formula gives .