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

• A

  $2.5 \mathrm{A}$
• Correct answer
  B

  $3.5 \mathrm{A}$
• C

  $4.0 \mathrm{A}$
• D

  zero

Solution

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=16 \mathrm{V}$
$\therefore 4i_1+2\left(i_1+i_2\right)-3+4i_1=16 \mathrm{V}$     ......(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
$i_1=1.5 \mathrm{A}$ and $i_2=2 \mathrm{A}$

$\therefore$ Current through $2 \mathrm{\Omega }$ resistor $=2+1.5=3.5 \mathrm{A}$

Hence, the correct answer is option: 'B'.