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 \[i_{2}^{2}\,t=0.25\,W\] |
\[\therefore \] \[H=12\times 0.25=3W\] |
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