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. $2.5A$
• B. $3.5A$
• C. $4.0A$
• 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 $=2+1.5=3.5A$