question_answer

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)  2.0                                        

B)  1.0

C)  0.5                                        

D)  3.0

Correct Answer: D

Solution :

                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,                  \[R\times {{i}_{1}}=8{{i}_{2}}\] or            \[4\times {{i}_{1}}=8{{i}_{2}}\] or            \[{{i}_{1}}=\frac{8}{4}{{i}_{2}}=2{{i}_{2}}\] or            \[{{i}_{1}}=2{{i}_{2}}\]                    ??..(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                 \[H=i_{1}^{2}(3)t\]                 \[={{(2{{i}_{2}})}^{2}}(3)t\]                 \[=12\,i_{2}^{2}\,t\] but         \[i_{2}^{2}\,t=0.25W\] \[\therefore \]  \[H=12\times 0.25=3W\]