With the given definitions of the voltages and currents on the circuit,we need
to apply KCL and KVL systematically in order to solve the problem. One obtains
• KVL(1 → 2 → 3 → 1): v_{x}+ v_{y} − 16 = 0 \longrightarrow  v_{x}+ v_{y} = 16V,
• KCL(2): i_{x} + 3 − i_{y}= 0\longrightarrow i_{x} − i_{y} = −3A.
Then, using Ohm’s law, v_{x} = 8i_{x} and v_{y} = 12i_{y}, leading to 2i_{x} + 3ii_{y}= 4. Finally, solving
for i_{x}, we obtain i_{x} = −1A, i_{y} = 2A, v_{y} = 24V, and
v_{x} = −8 V
as the voltage across the 8Ωresistor.As one verification of the solution, one can calculate
the powers of all components:
p_{8Ω} = v_{x}i_{x} = 8 W,
p_{12Ω} = v_{y}i_{y} = 48 W,
p_{16V} = 16 × (−i_{x}) = 16 W,
p_{3A} = (−v_{y}) × 3 = −72 W,
and check that
p_{8Ω} + p_{12Ω} + p_{16V}+ p_{3A} = 8 + 48 + 16 − 72 = 0.
As far as the power values are concerned, we can conclude that the current source delivers
power, while the resistors, as well as the voltage source, consume power.