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## Q. 1.9

Consider the following circuit involving a voltage source, which is connected to two resistors and a bulb B.Find the power of the bulb.

## Verified Solution

• In most circuits, the voltage and current directions are not defined a priori.
Therefore, when analyzing such a circuit, we define the directions arbitrarily, while
enforcing the sign convention for all components. When a current/voltage value
is found to be negative, we understand that the initial assumption is not correct.
However, this is not a problem at all, provided that we are consistent with the directions
throughout the solution.
For the circuit above, we label the nodes, define the directions of the currents, and
define the voltages in accordance with the sign convention, as follows.

Using KVL, one obtains
• KVL(1 → 2 → 3 → 1): $−10 + 3 × 2 + v_{x } = 0 \longrightarrow 𝑣x = 4V$.
Then, using Ohm’s law, $i_{y} = 𝑣_{x}∕4 = 1A$, and we further have
• KCL(2): 3 − $i_{x } − i_{y } = 0 \longrightarrow i_{x }$ = 3 − 1 = 2A.
Finally, the power of the device is found to be
$p_{B }$= $v_{x }$$i_{x }$= 4 × 2 = 8 W.