The power dissipated in the circuit shown in the figure is 30 Watt. The value of R is [AIPMT (M) 2012] |
A) \[20\,\,\Omega \]
B) \[15\,\,\Omega \]
C) \[10\,\,\Omega \]
D) \[30\,\,\Omega \]
Correct Answer: C
Solution :
[c] Here, \[{{R}_{1}}=R=?\] |
\[{{R}_{2}}=5\Omega ,V=10\,V\] |
and \[p=30\,W\] |
Hence \[p=\frac{{{V}_{2}}}{{{R}_{1}}}+\frac{{{V}_{2}}}{{{R}_{2}}}\] |
\[\frac{{{10}^{2}}}{R}=30-\frac{{{10}^{2}}}{5}\] |
\[\frac{100}{R}=30-20\] |
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