Question

In the part of the circuit shown in the figure, the potential difference between points $V_{A}-V_{B}=16V$ . The current passing through the $2\Omega$ resistance will be

• A. $2.5 \, A$
• B. $3.5 \, A$
• C. $4.0 \, A$
• D. zero

Answer 1

Answer: (B)

$V_{A}-V_{B}=16V$
$\therefore 4i_{1}+2\left(i_{1} + i_{2}\right)-3+4i_{1}=16V$ ......(i)
Using Kirchhoff's second law in the closed loops we have
$\Rightarrow 9-i_{2}-2\left(i_{1} + i_{2}\right)=0$ ......(ii)
Solving equations (i) and (ii), we get
$i_{1}=1.5A$ and $i_{2}=2A$
$\therefore $ Current through $2\Omega$ resistor $=2+1.5=3.5A$