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(C) Before connecting the cell $E$,the circuit consists of three resistors $6\,\Omega$,$8\,\Omega$,and $10\,\Omega$ in series with a $12\,V$ battery.T

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$4$

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(C) Before connecting the cell $E$,the circuit consists of three resistors $6\,\Omega$,$8\,\Omega$,and $10\,\Omega$ in series with a $12\,V$ battery.The total resistance $R_{eq} = 6 + 8 + 10 = 24\,\Omega$.The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{12}{24} = 0.5\,A$.After connecting the cell $E$ across the $8\,\Omega$ resistor,the current through the $8\,\Omega$ resistor remains the same $(0.5\,A)$.This implies that the potential difference across the $8\,\Omega$ resistor must remain the same as it was before,which is $V_{8\Omega} = I 	imes R = 0.5 	imes 8 = 4\,V$.Since the cell $E$ is connected in parallel with the $8\,\Omega$ resistor,the potential difference across the $8\,\Omega$ resistor is equal to the $EMF$ of the cell $E$.Therefore,$E = 4\,V$.

In the circuit shown,the current through the $8\,\Omega$ resistor is the same before and after connecting the cell $E$. The value of $E$ is .................... $V$.

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