Question 15: Current Electricity

Question

Two batteries, one of emf $$18 \mathrm{~V}$$ and internal resistance $$2 \Omega$$ and the other of emf $$12 \mathrm{~V}$$ and internal resistance $$1 \Omega$$ are connected as shown. The voltmeter $$V$$ will record a reading of

Accepted Answer

It is clear that the two cells oppose each other hence the effective emf in closed circuit is $$18-12=6 \mathrm{~V}$$ and net resistance is $$1+2=3 \Omega$$ (because in the closed circuit the internal resistance of two cells are in series)
The current in circuit will be in direction of arrow shown in figure
$$\begin{aligned}
I & =\frac{	ext { effective emf }}{	ext { total resistance }} \\
& =6 / 3=2 \mathrm{~A}
\end{aligned}$$
The potential difference across V will be
same as the terminal voltage of either cell
Since, current is drawn from the cell of $$18 \mathrm{~V}$$ hence
$$\begin{aligned}
V_1 & =E_1-I r_1=18-(2 	imes 2) \\
& =18-4=14 \mathrm{~V}
\end{aligned}$$

Answer

14 V

Answer

15 V

Answer

18 V

Answer

30 V

Current Electricity

Two batteries, one of emf $18 \mathrm{~V} and internal resistance 2 \Omega and the other of emf 12 \mathrm{~V} and internal resistance 1 \Omega are connected as shown. The voltmeter V$ will record a reading of

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