First slide

Electrical power

Question

Power dissipated across the $8 Ω$ resistor in the circuit shown here is 2 W. The power dissipated in watt units across the $3 Ω$ resistor is

Moderate

A

2.0
B

1.0
C

0.5
D

3.0

Solution

Resistances $1 Ω$ and $3 Ω$ are connected in series, so effective resistance
$R^{'} = 1 + 3 = 4 Ω$
Now, $R^{'}$ and $8 Ω$ are in parallel. We know that potential difference across resistances in parallel order is the same.
Hence, $R^{'} \times i_{1} = 8 i_{2}$
$or 4 \times i_{1} = 8 i_{2} or i_{1} = \frac{8}{4} i_{2} = 2 i_{2}$
or i₁ = 2i₂ …..(i)
Power dissipated across $8 Ω$ resistance is
$\begin{array}{l} i_{2}^{2} (8) t = 2 W \\ or i_{2}^{2} t = \frac{2}{8} = 0 .25 W \end{array}$ ….(ii)
Power dissipated across $3 Ω$ resistance is
$\begin{array}{l} H = i_{1}^{2} (3) t = {(2 i_{2})}^{2} (3) t = 12 i_{2}^{2} t \\ But i_{2}^{2} t = 0 .25 W \\ H = 12 \times 0 .25 = 3 W \end{array}$

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