Question

Find the emf and internal resistance of a single battery which equivalent to a combination of three batteries as shown in figure.

• A. $3V\text{and}1\Omega$
• B. $3V\text{and}2\Omega$
• C. $6V\text{and}1\Omega$
• D. $9V\text{and}1\Omega$

Answer 1

The correct option is B $3V\text{and}2\Omega$

The given combination consists of two batteries in parallel and the resultant of these two in series with the third one.

Given:
$E_1=10\text{V};E_2=4\text{V}{r}_1=2\Omega ;r_2=2\Omega$

For parallel combination we can apply,

$E_{eq}=\frac{\frac{E_1}{r_1}-\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}=\frac{\frac{10}{2}-\frac{4}{2}}{\frac{1}{2}+\frac{1}{2}}=3\text{V}$

Further equivalent internal resistance,

$\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}=\frac{1}{2}+\frac{1}{2}=1$

$\therefore r_{eq}=1\Omega$

Now this is in series with the third one, i.e.,


The equivalent emf of these two is $(6-3)\text{V}$ or $3\text{V}$ and equivalent internal resistance will be, $(1+1)$ or $2\Omega$.


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