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(A) The circuit consists of two cells in parallel with an external resistor of $4 \, \Omega$. First, we find the equivalent $EMF$ $(E_{eq})$ and equiv

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$8.4$

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(A) The circuit consists of two cells in parallel with an external resistor of $4 \, \Omega$. First, we find the equivalent $EMF$ $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ of the two parallel branches containing the cells. The first branch has $E_1 = 20 \, V$ and $r_1 = 4 \, \Omega$. The second branch has $E_2 = 6 \, V$ and $r_2 = 3 \, \Omega$. Using the formula for parallel combination of cells: $E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{20}{4} + \frac{6}{3}}{\frac{1}{4} + \frac{1}{3}} = \frac{5 + 2}{\frac{3+4}{12}} = \frac{7}{\frac{7}{12}} = 12 \, V$. $r_{eq} = \frac{1}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{1}{\frac{1}{4} + \frac{1}{3}} = \frac{1}{\frac{7}{12}} = \frac{12}{7} \, \Omega$. Now, this equivalent circuit is in parallel with an external resistor $R = 4 \, \Omega$. The terminal voltage $V$ across the parallel combination is given by: $V = E_{eq} 	imes \frac{R}{R + r_{eq}} = 12 	imes \frac{4}{4 + \frac{12}{7}} = 12 	imes \frac{4}{\frac{28+12}{7}} = 12 	imes \frac{4 	imes 7}{40} = 12 	imes \frac{28}{40} = 12 	imes 0.7 = 8.4 \, V$.

What will be the reading of the voltmeter in the given circuit (in $V$)?

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