First slide

Electrical power

Question

A current of 3 A flows through the $2 Ω$ resistor as shown in the circuit. The power dissipated in the $5 Ω$ resistor is

Moderate

A

4 W
B

2 W
C

1 W
D

5 W

Solution

Voltage across $2 Ω$ is the same as voltage across arm containing $1 Ω$ and $5 Ω$ resistances. Voltage across $2 Ω$ resistance,
$V = 2 \times 3 = 6 V$
So, voltage across lowest arm,
V₁ = 6 V
Current across $5 Ω, I = \frac{6}{1 + 5} = 1 A$
Thus, power across $5 Ω$
P = I² R = (1)² x 5 = 5 W

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