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## Q. 1.8

Consider the following circuit involving a voltage source, a current source, and two resistors.Find the voltage across the 8 Ω resistor.

## Verified Solution

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}$$i_{y}$ = −3A.
Then, using Ohm’s law, $v_{x}$ = 8$i_{x}$ and $v_{y}$ = 12$i_{y}$, leading to 2$i_{x}$ + 3i$i_{y}$= 4. Finally, solving
for $i_{x}$, we obtain $i_{x}$ = −1A, $i_{y}$ = 2A, $v_{y}$ = 24V, and
$v_{x}$ = −8 V
as the voltage across the 8Ωresistor.As one verification of the solution, one can calculate
the powers of all components:
$p_{8Ω}$ = $v_{x}$$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.