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:                                                                                              [AIPMT (S) 2006]

A)  2.0     

B)                   1.0      

C)  0.5                  

D)       3.0

Correct Answer: D

Solution :

[d] Resistance \[1\,\,\Omega \] 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 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         \[i_{2}^{2}\,t=\frac{2}{8}\,=0.25\,W\] (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        \[i_{2}^{2}\,t=0.25\,W\]

\[\therefore \]      \[H=12\times 0.25=3W\]