Question

Two identical batteries, each of emf $2\text{V}$ and internal resistance $r=1\Omega$ are connected as shown. The maximum power that can be developed across $R$ using these batteries is

• A. $3.2\text{W}$
• B. $8.2\text{W}$
• C. $2\text{W}$
• D. $4\text{W}$

Answer 1

The correct option is C $2\text{W}$

From the given diagram, we can easily observe that both cells are in parallel combination with same polarity. So, equivalent emf of cell is given by

$E_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}$

$\Rightarrow E_{eq}=\frac{\frac{2}{1}+\frac{2}{1}}{\frac{1}{1}+\frac{1}{1}}=2\text{V}$

And equivalent internal resistance will be

$r_{eq}=\frac{r_1{r}_2}{r_1+r_2}=\frac{1\times 1}{1+1}=0.5\Omega$

So, simplified circuit is given as


So, current in circuit will be
$i=\frac{E_{eq}}{R+r_{eq}}=\frac{2}{R+0.5}$

So, power developed through $R$ is given below

$P=i^{2}R=(\frac{2}{R+0.5}{)}^{2}R$

For, maximum power, $\frac{dP}{dR}=0$, we get

$R=0.5\Omega$

$\therefore P=(\frac{2}{0.5+0.5}{)}^{2}0.5=2\text{W}$

Hence, option $\left(c\right)$ is correct.

Alternate solution: $P=i^{2}R=(\frac{2}{R+0.5}{)}^{2}R$ From maximum power theorem, $R_{ext}=r_{int}$ $\therefore R=r_{eq}=\frac{r_1{r}_2}{r_1+r_2}=0.5\Omega$ So, $P=(\frac{2}{0.5+0.5}{)}^{2}0.5=2\text{W}$