A) 50 watt
B) 75 watt
C) 25 watt
D) 100 watt
Correct Answer: A
Solution :
The given current can be redrawn as follows: Here, \[r=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}=\frac{1.1}{1+1}=\frac{1}{2}\Omega \] And \[E=\frac{\sum{E/r}}{\sum{1/r}}=\frac{\left( \frac{10}{1} \right)+\left( \frac{10}{1} \right)}{1+1}=10\,V\] Maximum power will be developed across R when \[R=r=\frac{1}{2}\Omega \] \[\therefore \] \[{{i}_{\max }}=\frac{E}{r+R}=\frac{10}{\frac{1}{2}+\frac{1}{2}}=10A\] Hence, the correction option is [a].You need to login to perform this action.
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