• # question_answer Maximum power developed across variable resistance R in the circuit shown in the figure is A)  50 watt                      B)  75 wattC)  25 watt                      D)  100 watt

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].