A) 25
B) 20
C) 30
D) 35
E) 40
Correct Answer: C
Solution :
The simplified circuit is shown below: From figure, \[15{{I}_{1}}=5{{I}_{2}}\]or \[{{I}_{2}}=\frac{15{{I}_{1}}}{5}=3{{I}_{1}}\] \[\therefore \] \[I={{I}_{1}}+{{I}_{2}}=\frac{{{I}_{2}}}{3}+{{I}_{2}}=\frac{4{{I}_{2}}}{3}\] ...(i) But \[I_{2}^{2}\times 5=42\] \[I_{2}^{2}=\frac{42}{5}=8.4\] Putting value of\[{{I}_{2}}\]in Eq. (i), we get \[I=\frac{4}{3}\times \sqrt{8.4}\] Therefore, heat dissipated across \[2\,\Omega \] \[={{I}^{2}}\times 2\] \[=\frac{16}{9}\times 8.4\times 2\] \[\approx 30\,J{{s}^{-1}}\]You need to login to perform this action.
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