question_answer

Power dissipated across the \[8\,\Omega \] resistor in the circuit shown here is 2 W. 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 \[8\,\Omega \] 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 \[{{i}_{1}}=2{{i}_{2}}\] Power dissipated across \[8\Omega \] resistance is \[i_{2}^{2}(8)t=2W\] or \[i_{2}^{2}t=\frac{2}{8}=0.25\,W\]                          (ii) Power dissipated across \[3\Omega \] resistance is \[H=i_{1}^{2}(3)t={{(2{{i}_{2}})}^{2}}(3)t=12i_{2}^{2}t\] But \[i_{2}^{2}t=0.25\,W\] \[H=12\times 0.25=3\,W\]