A circuit consists of three batteries of emf | vedclass

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(B) The batteries are connected in parallel. The equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ of the combination are given by:$

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$2.0$

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(B) The batteries are connected in parallel. The equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ of the combination are given by:$E_{eq} = \frac{\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} + \frac{E_{3}}{r_{3}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}}$$\frac{1}{r_{eq}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}$Given $E_{1} = 1 	ext{ V}, r_{1} = 1 \Omega$; $E_{2} = 2 	ext{ V}, r_{2} = 2 \Omega$; $E_{3} = 3 	ext{ V}, r_{3} = 1 \Omega$.First, calculate the equivalent internal resistance:$\frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{1} = 1 + 0.5 + 1 = 2.5 \Omega^{-1} = \frac{5}{2} \Omega^{-1}$$r_{eq} = \frac{2}{5} \Omega = 0.4 \Omega$Now, calculate the equivalent emf:$E_{eq} = r_{eq} 	imes \left( \frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} + \frac{E_{3}}{r_{3}} ight)$$E_{eq} = 0.4 	imes \left( \frac{1}{1} + \frac{2}{2} + \frac{3}{1} ight) = 0.4 	imes (1 + 1 + 3) = 0.4 	imes 5 = 2.0 	ext{ V}$Thus, the potential difference between points $P$ and $Q$ is $2.0 	ext{ V}$.

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