Two batteries, one of e.m.f. $12 \mathrm{~V}$ and internal resistance $2 \Omega$ and other of e.m.f. $6…
Two batteries, one of e.m.f. $12 \mathrm{~V}$ and internal resistance $2 \Omega$ and other of e.m.f. $6 \mathrm{~V}$ and internal resistance $1 \Omega$, are connected as shown in the figure. What will be the reading of the voltmeter 'V'?
$12 \mathrm{~V}$
$8 \mathrm{~V}$
$6 \mathrm{~V}$
$4 \mathrm{~V}$
Solution
The formula for the equivalent emf of the parallel combination of batteries is
$\varepsilon_e=r_{\text {eq }}\left(\frac{e_1}{r_1}+\frac{c_2}{r_2}\right)$
Here, $r_{\mathrm{rq}}$ is the equivalent resistance
$\begin{aligned}
& \frac{1}{r_{\text {eq }}}=\frac{1}{r_1}+\frac{1}{r_2} \\
& \frac{1}{r_{\text {84 }}}=\frac{1}{2}+\frac{1}{1} \\
& \frac{1}{r_{\text {eq }}}=\frac{3}{2}
\end{aligned}$
Substituting the values
$\begin{aligned}
\varepsilon_e & =r_{\infty}\left(\frac{e_1}{r_1}+\frac{e_2}{r_2}\right) \\
\varepsilon_{\mathrm{e}} & =\frac{2}{3}\left(\frac{12}{2}+\frac{6}{1}\right) \\
\varepsilon_e & =\frac{2}{3} \times 12 \\
\therefore \quad \varepsilon_e & =8 \mathrm{~V}
\end{aligned}$
