Question

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

• A. $0A$
• B. $0.13A$, from $Q$ to $P$
• C. $0.13A$, from $P$ to $Q$
• D. $1.3A$, from $P$ to $Q$

Answer 1

The correct option is B $0.13A$, from $Q$ to $P$

$\text{Given:-}$ $R_1=2\Omega$

$R_2=1\Omega$

$R_3=3\Omega$

$R_4=3\Omega$

$V_1=6V$

$V_2=9V$

$\text{Solution:-}$

Connect point Q to the ground and apply KCL. Consider the grounded circuit as shown below.

Apply KCL of point Q we can write incoming current at Q $=$ outgoing current from Q

$\Rightarrow \frac{V+6}{3}+\frac{V}{1}=\frac{9-V}{5}$

$\Rightarrow V[\frac{1}{3}+\frac{1}{5}+1]=\frac{9}{5}-2$

$\Rightarrow V\left[\frac{5+3+15}{15}\right]=\frac{9-10}{5}$

$\Rightarrow V\left[\frac{23}{15}\right]=\frac{-1}{5}$

$\Rightarrow V=\frac{-3}{23}=-0.13V$

Thus, current in the 1Ω resistance is 0.13 A, from Q to P.

$\text{Hence the correct option is B}$