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

• Correct answer
  A

  3.0
• B

  2.0
• C

  1.0
• D

  0.5

Solution

$1 \Omega$ and $3 \Omega$ are in series so, $egin{aligned} & R_s=R_1+R_2 \\ & R_s=3+1=4 \Omega \end{aligned}$ and $R_2=8 \Omega$ net $i=$ current in the circuit Current through $R_1$ is $i_1=\frac{i \times R_2}{R_1+R_2}$ $=\frac{i \times 8}{12}=\frac{21}{3}$ Current $R_2$ is $i_2=\frac{1 \times R_1}{R_1+R_2}=\frac{i \times 4}{12}$ $=\frac{i \times 8}{12}$ Power dissipated in $3 \Omega$ resistance is: $P_1=l_1^2 \times 3$ Power dissipated in $8 \Omega$ resistance is: $P_2=l_2^2 \times 8$ For (i) and (ii) $\begin{aligned} & \frac{P_1}{P_2}=\frac{i_1^2 \times 3}{i_2^2 \times 8} \\ & \Rightarrow \frac{P_1}{P_2} \end{aligned}$ $\begin{aligned} & =\frac{\left(\frac{2 i}{3}\right)^2 \times 3}{\left(\frac{i}{3}\right)^2 \times 8}=\frac{3}{2} \\ \therefore \quad P_1 & =\frac{3}{2} \times P_2=\frac{3}{2} \times 2 =3 \mathrm{~W} \end{aligned}$ $\begin{aligned} \because & P & =100 \mathrm{~W}, V=125 V \\ \therefore & P & =V I \\ \Rightarrow & I & =\frac{P}{V}=\frac{100}{125} A \end{aligned}$

Hence, the correct answer is option: 'A'.