Question 195 of 380

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

A$0.13A$, from $P$ to $Q$B$1.3A$, from $P$ to $Q$C$1.3A$, from $Q$ to $P$D$0.13A$, from $Q$ to $P$

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Answer: (D) $0.13A$, from $Q$ to $P$

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

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