Question 560: Current Electricity

Question

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'?

Accepted Answer

The formula for the equivalent emf of the parallel combination of batteries is
$$\varepsilon_{\mathrm{e}}=r_{	ext {eq }}\left(\frac{e_1}{r_1}+\frac{e_2}{r_2}ight)$$
Here, $$r_{	ext {eq }}$$ is the equivalent resistance
$$\begin{aligned}
& \frac{1}{r_{	ext {eq }}}=\frac{1}{r_1}+\frac{1}{r_2} \
& \frac{1}{r_{	ext {eq }}}=\frac{1}{2}+\frac{1}{1} \
& \frac{1}{r_{	ext {eq }}}=\frac{3}{2}
\end{aligned}$$
Substituting the values
$$\begin{aligned}
\varepsilon_{\mathrm{e}} & =\mathrm{r}_{\mathrm{eq}}\left(\frac{\mathrm{e}_1}{\mathrm{r}_1}+\frac{\mathrm{e}_2}{\mathrm{r}_2}ight) \
\varepsilon_{\mathrm{c}} & =\frac{2}{3}\left(\frac{12}{2}+\frac{6}{1}ight) \
\varepsilon_{\mathrm{e}} & =\frac{2}{3} 	imes 12 \
	herefore \quad \varepsilon_{\mathrm{e}} & =8 \mathrm{~V}
\end{aligned}$$

Answer

12 V

Answer

8 V

Answer

6 V

Answer

4 V

Current Electricity

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'?

Choose the correct answer: