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(B) The resistors $4\,\Omega$ and $3\,\Omega$ are in parallel. Since the potential difference across parallel branches is the same,the current $i_1$ t

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$3.2$

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(B) The resistors $4\,\Omega$ and $3\,\Omega$ are in parallel. Since the potential difference across parallel branches is the same,the current $i_1$ through the $3\,\Omega$ resistor is given by $4\,\Omega 	imes 1\,A = 3\,\Omega 	imes i_1$,which gives $i_1 = \frac{4}{3}\,A$.The total current flowing through the upper branch is $I_{upper} = 1\,A + \frac{4}{3}\,A = \frac{7}{3}\,A$.The equivalent resistance of the $4\,\Omega$ and $3\,\Omega$ parallel combination is $R_{eq} = \frac{4 	imes 3}{4 + 3} = \frac{12}{7}\,\Omega$.The potential difference between points $P$ and $M$ is $V_{PM} = I_{upper} 	imes R_{eq} = \frac{7}{3}\,A 	imes \frac{12}{7}\,\Omega = 4\,V$.The lower branch consists of two $0.5\,\Omega$ resistors in parallel,which is equivalent to $R_{lower1} = \frac{0.5 	imes 0.5}{0.5 + 0.5} = 0.25\,\Omega$. This is in series with a $1\,\Omega$ resistor. The total resistance of the lower branch is $R_{lower} = 0.25\,\Omega + 1\,\Omega = 1.25\,\Omega$.The current in the lower branch is $I_{lower} = \frac{V_{PM}}{R_{lower}} = \frac{4\,V}{1.25\,\Omega} = 3.2\,A$.The potential difference between points $M$ and $N$ is the voltage drop across the $1\,\Omega$ resistor: $V_{MN} = I_{lower} 	imes 1\,\Omega = 3.2\,A 	imes 1\,\Omega = 3.2\,V$.

In the circuit shown,the current through the $4\,\Omega$ resistor is $1\,A$ when the points $P$ and $M$ are connected to a d.c. voltage source. The potential difference between the points $M$ and $N$ is ............... $V$.

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