MCQ
In the circuit shown in figure potential difference between points A and $B$ is $16\,V$. the current passing through $2 \Omega$ resistance will be $...........\,A$


- A$2.5$
- ✓$3.5$
- C$4.0$
- D$0$

$\therefore 4 i _1+2\left( i _1+ i _2\right)-3+4 i _1=16\, V$
Using Kirchhoff's second law in the closed loop we have
$9-i_2-2\left(i_1+i_2\right)=0$
Solving equations $(i)$ and $(ii)$, we get $i _1=1.5 A ^{\text {and } i _2}=2 A$
$\therefore$ current through $2 W$ resistor $=2+1.5=3.5\,A$.
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$\left\lfloor{m}_{e}=\text { mass of electron }=9 \times 10^{-31} \,{kg}\right.$
${h}=\text { Planck constant }=6.6 \times 10^{-34} {Js}$
$\left.{k}_{{B}}=\text { Boltzmann constant }=1.38 \times 10^{-23}\, {JK}^{-1}\right]$

