Question

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

• A. 2
• B. 1
• C. 0.5
• D. 3

Answer 1

Answer: (D)

Resistances $1\Omega$ and $3\Omega$ are connected in series,
so effective resistance $R=1+3=4\Omega$
Now, $R$ and $80$ are in parallel.
We know that potential difference across resistances in parallel order is same
Hence, $R\times i_1=8i_2$
or $4\times i_1=8i_2$
or $i_1=\frac{8}{4}{i}_2=2i_2$
or $i_1=2i_2$ ...(i)
Power dissipated across $8\Omega$ resistance is
$i_2^2(8)t=2W$
or $i_2^{2}t=\frac{2}{8}=0.25W$ ...(ii)
Power dissipated across $3\Omega$ resistance is
$H=i_1^2(3)t={\left(2i_2\right)}^2(3)t$
$=12i_2^{2}t$
but $i_2^{2}t=0.25W$
$\therefore H=12\times 0.25=3W$