Question

For the circuit shown in figure, match the following columns:

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \left(A\right)\text{Current in wire ae}& \left(p\right)1A\\ \left(B\right)\text{Current in wire be}& \left(q\right)2A\\ \left(C\right)\text{Current in wire ce}& \left(r\right)0.5A\\ \left(D\right)\text{Current in wire de}& \left(s\right)\text{none of these}\end{array}$

• A. $\left(A\right)\to \left(s\right),\left(B\right)\to \left(s\right),\left(C\right)\to \left(q\right),\left(D\right)\to \left(p\right)$
• B. $\left(A\right)\to \left(r\right),\left(B\right)\to \left(p\right),\left(C\right)\to \left(q\right),\left(D\right)\to \left(s\right)$
• C. $\left(A\right)\to \left(q\right),\left(B\right)\to \left(s\right),\left(C\right)\to \left(q\right),\left(D\right)\to \left(s\right)$
• D. $\left(A\right)\to \left(p\right),\left(B\right)\to \left(q\right),\left(C\right)\to \left(r\right),\left(D\right)\to \left(s\right)$

Answer 1

The correct option is C $\left(A\right)\to \left(q\right),\left(B\right)\to \left(s\right),\left(C\right)\to \left(q\right),\left(D\right)\to \left(s\right)$

Let potential of point $e$ be $V$ volt.
The current through any wire can be obtained from ohm's law, $I=\frac{\Delta V}{R}$
Using Kirchhoffs junction law,
$I_{ae}+I_{be}+I_{ce}+I_{de}=0$
$\Rightarrow \left(\frac{2-V}{1}\right)+\left(\frac{4-V}{2}\right)+\left(\frac{6-V}{1}\right)+\left(\frac{4-V}{2}\right)=0$
$\Rightarrow V=4$Volt

Individual values of current are
$I_{ae}=\frac{2-V}{1}=\frac{2-4}{1}=-2A$
$I_{be}=\frac{4-V}{2}=\frac{4-4}{2}=0$
$I_{ce}=\frac{6-V}{1}=\frac{6-4}{1}=2A$
$I_{de}=\frac{4-V}{2}=\frac{4-4}{2}=0$

As magnitudes of current is asked, correct options are
$\left(A\right)\to \left(q\right),\left(B\right)\to \left(s\right),\left(C\right)\to \left(q\right),\left(D\right)\to \left(s\right)$