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

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

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(C) When two cells with $emf$ $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ 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 the 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$The voltmeter connected across the parallel combination 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 in the figure. The voltmeter $V$ will record a reading of ............. $V$.

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