Two batteries,one of emf 18 , V and internal | vedclass

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(C) When two cells with emf $E_1, E_2$ and internal resistances $r_1, r_2$ are connected in parallel,the equivalent emf $E_{eq}$ is given by the formu

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$14$

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(C) When two cells with emf $E_1, E_2$ and internal resistances $r_1, r_2$ are connected in parallel,the equivalent emf $E_{eq}$ is given by the formula:$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$Given:$E_1 = 18 \, V, r_1 = 2 \, \Omega$$E_2 = 12 \, V, r_2 = 1 \, \Omega$Substituting these values into the formula:$E_{eq} = \frac{(18 	imes 1) + (12 	imes 2)}{2 + 1}$$E_{eq} = \frac{18 + 24}{3}$$E_{eq} = \frac{42}{3} = 14 \, V$Since the voltmeter is connected across the parallel combination,it will measure the equivalent emf of the circuit.Therefore,the reading of the voltmeter is $14 \, V$.

Two batteries,one of emf $18 \, V$ and internal resistance $2 \, \Omega$ and the other of emf $12 \, V$ and internal resistance $1 \, \Omega$,are connected in parallel as shown. The voltmeter $V$ will record a reading of ............. $V$.

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