Two batteries,one of e.m.f. 12 text{ V} and i | vedclass

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(B) The formula for the equivalent e.m.f. of the parallel combination of batteries is given by $\varepsilon_{eq} = r_{eq} \left( \frac{\varepsilon_1}{

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$8$

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(B) The formula for the equivalent e.m.f. of the parallel combination of batteries is given by $\varepsilon_{eq} = r_{eq} \left( \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} ight)$.First,we calculate the equivalent internal resistance $r_{eq}$:$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{2} + \frac{1}{1} = \frac{3}{2} \Omega^{-1}$.Therefore,$r_{eq} = \frac{2}{3} \Omega$.Now,substitute the values into the formula for equivalent e.m.f.:$\varepsilon_{eq} = \frac{2}{3} \left( \frac{12}{2} + \frac{6}{1} ight) = \frac{2}{3} (6 + 6) = \frac{2}{3} 	imes 12 = 8 	ext{ V}$.The voltmeter measures the potential difference across the parallel combination,which is equal to the equivalent e.m.f. of the circuit.Thus,the reading of the voltmeter is $8 	ext{ V}$.

Two batteries,one of e.m.f. $12 \text{ V}$ and internal resistance $2 \Omega$ and other of e.m.f. $6 \text{ V}$ and internal resistance $1 \Omega$,are connected as shown in the figure. What will be the reading of the voltmeter '$V$' (in $\text{ V}$)?

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