In the circuit shown, the current through the $4\,\Omega $ resistor is $1\, amp$ when the points $P$ and $M$ are connected to a d.c. voltage source. The potential difference between the points $M$ and $N$ is ............... $volt$
AIPMT 2008, Medium
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As the P.D.
between $4 \Omega$ and $3 \Omega$
(in parallel), arc the same, $4 \times 1$ amp $=3 \times i_{1} \Rightarrow i_{1}=\frac{4}{3} A$
Total resistance of $4 \Omega$ and $3 \Omega=12 / 7 \Omega$
Current in $M Q P$ (upper one) $=1+\frac{4}{3}=\frac{7}{3} A$ $\therefore \quad P . D .=\frac{12}{7} \times \frac{7}{3}=4 V$
Current in $M N P=\frac{4}{1.25}=\frac{4 \times 4}{5}=\frac{16}{5} A$
$\therefore \quad$ P.D. across $1 \Omega=\frac{16}{5} A \times 1 \Omega=\frac{16}{5}$ volt
$\Rightarrow$ P.D. across $1 \Omega=3.2$ volt.
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