A battery of emf 18 , V and internal resistan | vedclass

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(B) The equivalent emf $(E_{eq})$ of two batteries in parallel is given by $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$.Here, $E_1 = 18 \, V$, $

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$12$

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(B) The equivalent emf $(E_{eq})$ of two batteries in parallel is given by $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$.Here, $E_1 = 18 \, V$, $r_1 = 3 \, \Omega$, $E_2 = 10 \, V$, and $r_2 = 1 \, \Omega$.Since the batteries are connected with opposite polarities (as seen in the figure), we take $E_2 = -10 \, V$.$E_{eq} = \frac{18/3 + (-10)/1}{1/3 + 1/1} = \frac{6 - 10}{4/3} = \frac{-4}{4/3} = -3 \, V$.The equivalent internal resistance is $r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{3 	imes 1}{3 + 1} = 0.75 \, \Omega$.The potential difference across the parallel combination is the terminal voltage $V = |E_{eq}| = 3 \, V$. However, looking at the circuit as a loop, the current $i$ flows from the $18 \, V$ battery to the $10 \, V$ battery: $i = \frac{18 - 10}{3 + 1} = \frac{8}{4} = 2 \, A$.The potential difference across the $18 \, V$ battery is $V = E_1 - i r_1 = 18 - (2 	imes 3) = 18 - 6 = 12 \, V$.The potential difference across the $10 \, V$ battery is $V = E_2 + i r_2 = 10 + (2 	imes 1) = 12 \, V$.Thus, the voltmeter reading is $12 \, V$.

$A$ battery of emf $18 \, V$ and internal resistance $3 \, \Omega$ and another battery of emf $10 \, V$ and internal resistance $1 \, \Omega$ are connected in parallel as shown in the figure. The voltmeter reading is: (in $ \, V$)

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