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(D) For two cells with emf $\varepsilon_1$ and $\varepsilon_2$ and internal resistances $r_1$ and $r_2$ connected in parallel,the equivalent emf $\var

Vedclass · Accepted Answer

$2.67$

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(D) For two cells with emf $\varepsilon_1$ and $\varepsilon_2$ and internal resistances $r_1$ and $r_2$ connected in parallel,the equivalent emf $\varepsilon_{eq}$ is given by the formula:$\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$Given values:$\varepsilon_1 = 2 \ V, r_1 = 0.1 \ \Omega$$\varepsilon_2 = 4 \ V, r_2 = 0.2 \ \Omega$Substituting these values into the formula:$\varepsilon_{eq} = \frac{(2 	imes 0.2) + (4 	imes 0.1)}{0.1 + 0.2}$$\varepsilon_{eq} = \frac{0.4 + 0.4}{0.3}$$\varepsilon_{eq} = \frac{0.8}{0.3} \approx 2.67 \ V$Thus,the correct option is $D$.

One electric cell (having emf of $2 \ V$ and internal resistance of $0.1 \ \Omega$) and another electric cell (having emf of $4 \ V$ and internal resistance of $0.2 \ \Omega$) are connected in parallel to each other. The equivalent emf will be . . . . . . $V$.

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