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(D) The circuit consists of two parallel branches connected in series with a $1\,\Omega$ resistor and the cell.First,calculate the equivalent resistan

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(D) The circuit consists of two parallel branches connected in series with a $1\,\Omega$ resistor and the cell.First,calculate the equivalent resistance of the two parallel branches.The upper branch has resistances $4\,\Omega$ and $2\,\Omega$ in series,so $R_{upper} = 4 + 2 = 6\,\Omega$.The lower branch has resistances $2\,\Omega$ and $4\,\Omega$ in series,so $R_{lower} = 2 + 4 = 6\,\Omega$.The equivalent resistance of these two parallel branches is $R_p = \frac{6 	imes 6}{6 + 6} = 3\,\Omega$.The total external resistance is $R_{ext} = 3\,\Omega + 1\,\Omega = 4\,\Omega$.The total current $i$ from the cell is $i = \frac{E}{R_{ext} + r} = \frac{10}{4 + 1} = 2\,A$.This current splits equally into the two parallel branches because both branches have a total resistance of $6\,\Omega$. Thus,the current in the upper branch is $i_1 = 1\,A$ and in the lower branch is $i_2 = 1\,A$.Now,calculate the potentials at $A$ and $B$ relative to the negative terminal of the cell (let its potential be $0\,V$):$V_A = E - i_1 	imes 4 = 10 - 1 	imes 4 = 6\,V$.$V_B = E - i_2 	imes 2 = 10 - 1 	imes 2 = 8\,V$.Therefore,$V_A - V_B = 6 - 8 = -2\,V$.

In the circuit shown below,the cell has an $e.m.f.$ of $10\,V$ and internal resistance of $1\,\Omega$. The other resistances are shown in the figure. The potential difference ${V_A} - {V_B}$ is ................ $V$.

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