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(B) The circuit consists of two batteries in a loop. The equivalent electromotive force $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ can b

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$5$

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(B) The circuit consists of two batteries in a loop. The equivalent electromotive force $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ can be calculated using the formula for parallel combination of cells:$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$Given $E_1 = 6 	ext{ V}$, $r_1 = 1 	ext{ }\Omega$, $E_2 = 3 	ext{ V}$, $r_2 = 2 	ext{ }\Omega$ (note the polarity, the $3 	ext{ V}$ battery is connected in opposition to the $6 	ext{ V}$ battery).$E_{eq} = \frac{\frac{6}{1} - \frac{3}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{6 - 1.5}{1.5} = \frac{4.5}{1.5} = 3 	ext{ V}$.Alternatively, using Kirchhoff's loop rule for the loop:$I = \frac{6 - 3}{1 + 2} = \frac{3}{3} = 1 	ext{ A}$.The potential difference across $AB$ is $V_{AB} = E_1 - I r_1 = 6 - (1 	imes 1) = 5 	ext{ V}$.Checking with the second branch: $V_{AB} = E_2 + I r_2 = 3 + (1 	imes 2) = 5 	ext{ V}$.Thus, the voltage across $AB$ is $5 	ext{ V}$.

Two batteries of different emfs and different internal resistances are connected as shown. The voltage across $AB$ in volts is

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