In the circuit shown, the current in the $1 \mathrm{\Omega }$ resistor is:

• A

  $0.13 \mathrm{A}$, from $P$ to $Q$
• B

  $1.3 \mathrm{A}$, from $P$ to $Q$
• C

  $1.3 \mathrm{A}$, from $Q$ to $P$
• Correct answer
  D

  $0.13 \mathrm{A}$, from $Q$ to $P$

Solution

$R=3 \mathrm{\Omega }+2 \mathrm{\Omega }$. Let potential at $P=0$, $\therefore V_A=6 \mathrm{V},V_D=-9 \mathrm{V}$, and$VQ=x$. Apply K.C.L. at $x$, $\frac{x-6}{3}+\frac{x-0}{1}+\frac{x+9}{5}=0$. Solving, $x=\frac{3}{23 } \mathrm{V}$. So, current in $1 \mathrm{\Omega }$$\Rightarrow i=\frac{\frac{3}{23}-0}{1}$ from $Q$ to $P$$\Rightarrow i=0.13 \mathrm{A}$ from $Q$ to $P$

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