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'?

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
\(\begin{aligned}
\varepsilon_e & =r_{\infty}\left(\frac{e_1}{r_1}+\frac{e_2}{r_2}\right) \\
\varepsilon_{\mathrm{e}} & =\frac{2}{3}\left(\frac{12}{2}+\frac{6}{1}\right) \\
\varepsilon_e & =\frac{2}{3} \times 12 \\
\therefore \quad \varepsilon_e & =8 \mathrm{~V}
\end{aligned}\)