a) Use superposition to find the voltage v.
b) Find the power dissipated in the 10 ohm resistor.

First "kill" the current source.

I will use circuit reduction to find v, being careful not to use a transform or resistor combination that will cause v to get lost.

Transform the 110 volt source and the 5 ohm resistor, and combine the 2 ohm and 12 ohm resistors in series.

Circuit A

Combine the 5 ohm and 14 ohm resistors in parallel

The current I through the 10 ohm resistor can now be obtained using current division

I = 22 [3.684 / (3.684 + 10)] = 5.923 A

Thus the voltage across the 10 ohm resistor is

v = 10 I
v = 59.23 V (due to 110 volt source)

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NOTE: This result could be obtained a bit more directly from Circuit A above by combining all three resistors in parallel, then calculating the voltage developed across that resistor with 22 amps going through it.

R_eq = 5 || 10 || 14 = 2.692 W

v = 22 (2.692) = 59.23 V

This is valid since the two nodes across which we want the voltage are unaffected by the combination. This does, however, involve combining the resistor where v appears with others. In the introductory course, messing with components containing parameters of interest is generally discouraged, because those parameters can easily become lost. With experience, you will develop a better feel for when this works and when it does not.

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Now put the current source back in the circuit and "kill" the voltage source.

Although v could be obtained in this circuit using resistor combinations, it would be very easy to lose v, so I will use mesh analysis instead.

Define current source in terms of mesh currents.

I_A = - 4

Removing the current source leaves only meshes B and C. Write KVL around the two remaining meshes.

B:

5 (I_B - I_A) + 10 (I_B - I_C) = 0
- 5 I_A + 15 I_B - 10 I_C = 0
- 5 (- 4) + 15 I_B - 10 I_C = 0
15 I_B - 10 I_C = - 20 (equation 1)

C:

2 (I_C - I_A) + 12 I_C + 10 (I_C - I_B) = 0
- 2 I_A - 10 I_B + 24 I_C = 0
- 2 (- 4) - 10 I_B + 24 I_C = 0
- 10 I_B + 24 I_C = - 8 (equation 2)

Solving the two equations gives

I_B = - 2.154
I_C = - 1.231

Calculating v:

v = 10 (I_B - I_C) = 10 (- 2.154 - (- 1.231))
v = - 9.23 (due to 4 A source)

v = 59.23 - 9.23
v = 50 V

P = v² / R
P = 50² / 10
P = 250 W