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 80 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 \[i_{2}^{2}t=\frac{2}{8}=0.25W\] ...(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 \[i_{2}^{2}\,t=0.25W\] \[\therefore \] \[H=12\times 0.25=3W\]You need to login to perform this action.
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