a) In the circuit below find i₁, i₂, and i₃, using nodal analysis. Note that the items in Blue were not on the original diagram but were added to assist in the solution.

I have chosen the bottom node as the reference node and labeled the other two as V_A and V_B.

The 80 V source directly connects node B to the reference node, thus

V_B = 80

and node B becomes part of a supernode including the reference node, so no KCL equation will be required for that supernode.

We still must write KCL at node A. I will sum the currents out of the node.

First we need to determine V₅ in order to determine the current out of node A through the 5K resistor. Equating the difference in node voltages with the summ of voltages along the left-most branch of the circuit yields

V_A = V₅ - 30

so

V₅ = V_A + 30

Now we can write KCL at node A.

and substituting for V_B and V₅ gives

Multiplying through by 5000 gives

V_A + 30 + 10 V_A + 5 V_A - 400 + 50 = 0
16 V_A = 320
V_A = 20 volts

Now we can find the three desired currents.

We can get i₁ and i₂ using Ohm's Law

i₁ = - V₅ / 5K (note that the arrow enters the negative reference, thus the negative sign)
i₁ = - (V_A + 30) / 5K = - 50 / 5K
i₁ = - 10 mA

i₂ = V_A / 500 = 20 / 500
i₂ = 40 mA

For i₃, KCL at the bottom node gives

i₁ = i₂ + i₃ + i₄

thus

i₃ = i₁ - i₂ - i₄
i₃ = - 10m - 40m - V_B / 4K = - 50m - 80 / 4K
i₃ = - 70m

First calculate the power developed by the sources, making certain that none are actually absorbing power.

30 V source (note that the equation for power developed is negative since the arrow enters the positive - one of the few cases where a negative appears in this situation.)

P_dev = - 30 (i₁) = -30 (-10m) = 300 mW

10 mA source

P_dev = 10m (V_B - V_A) = 10m (80 - 20) = 600 mW

80 V source (See note on 30 V source calculation)

P_dev = - 80 (i₃) = - 80 (- 70m) = 5.6 W

So the total power developed is

P_Tdev = 300m + 600m + 5.6
P_Tdev = 6.5 W

Now calculate the power absorbed by the resistors.

5K (note that this equation has a negative sign inserted since the arrow enters the negative reference)

P_abs = - i₁ (V₅) = - (- 10m) (50) = 500 mW

500 W

P_abs = i₂ (V_A) = 40m (20) = 800 mW

1K (since we do not already have the current through the 1K resistor, I will calculate the power absorbed using P = V² / R)

P_abs = (V_B - V_A)² / 1K = 60² / 1K = 3.6 W

4K (see note on 1K above)

P_abs = (V_B)² / 4K = 80² / 4K = 1.6 W

Thus the total power absorbed (or dissipated) is

P_Tabs = 500m + 800m + 3.6 + 1.6
P_Tabs = 6.5 W

Since the total power developed equals the total power absorbed, the calculated values for the various voltages and currents are almost certainly correct.