A battery of emf E_0 = 12, V is connected acr | vedclass

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(C) The equivalent $emf$ $(E_{eq})$ of the two cells connected in parallel is given by the formula:$E_{eq} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r

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$25\,cm$

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(C) The equivalent $emf$ $(E_{eq})$ of the two cells connected in parallel is given by the formula:$E_{eq} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$Given $\varepsilon_1 = 2\,V, r_1 = 2\,\Omega$ and $\varepsilon_2 = 4\,V, r_2 = 6\,\Omega$:$E_{eq} = \frac{2 	imes 6 + 4 	imes 2}{2 + 6} = \frac{12 + 8}{8} = \frac{20}{8} = 2.5\,V$The total resistance of the potentiometer wire is $R_{wire} = 4\,m 	imes 4\,\Omega/m = 16\,\Omega$.The total resistance of the primary circuit is $R_{total} = R + R_{wire} = 8\,\Omega + 16\,\Omega = 24\,\Omega$.The current in the primary circuit is $I = \frac{E_0}{R_{total}} = \frac{12\,V}{24\,\Omega} = 0.5\,A$.The potential drop per unit length (potential gradient) is $x = I 	imes (resistance\,per\,unit\,length) = 0.5\,A 	imes 4\,\Omega/m = 2\,V/m$.At the balancing point $N$,the potential drop across length $l$ is equal to $E_{eq}$:$x 	imes l = E_{eq}$$2\,V/m 	imes l = 2.5\,V$$l = \frac{2.5}{2} = 1.25\,m = 125\,cm$.*Correction*: Re-evaluating the parallel combination based on the diagram polarity: $\varepsilon_1$ and $\varepsilon_2$ are opposing. $E_{eq} = \frac{\varepsilon_1/r_1 - \varepsilon_2/r_2}{1/r_1 + 1/r_2} = \frac{2/2 - 4/6}{1/2 + 1/6} = \frac{1 - 2/3}{4/6} = \frac{1/3}{2/3} = 0.5\,V$.Now,$2\,V/m 	imes l = 0.5\,V \implies l = 0.25\,m = 25\,cm$.

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