Question

In the circuit shown, the current through the 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:

• A. 0.5 volt
• B. 3.2 volt
• C. 1.5 volt
• D. 1.0 volt

Answer 1

Answer: (B)

3.2 volt

$i_1=$amp $=$ current through $4\Omega$
$i_2=$ current through $3\Omega$ and
$\frac{i_1}{i_2}=\frac{3}{4}\Rightarrow i_2=\frac{4}{3}amp$
$i_3=i_1+I_2=I+\frac{4}{3}=\frac{7}{3}amp$
and $\frac{i_3}{i_4}=\frac{3/4}{12/7}\Rightarrow \frac{7}{3}\times \frac{12}{7}=i_4\times \frac{5}{4}$
$\therefore i_4=\frac{16}{5}amp=3.2amp$
$V_{NM}=P.D.acrossNandM=i_4\times 1\Omega$
$=3.2\times 1=3.2volt$
Second approach.
$V_{PM}=1A\times 4\Omega =4volt$
$\frac{V_{PN}}{V_{NM}}=\frac{1}{4}and{V}_{PN}+V_{NM}=4V$
$\frac{1}{4}.V_{NM}+V_{NM}=4$
$\therefore V_{NM}=\frac{16}{5}V=3.2V$