Question

What is the current flowing through the $10\Omega$ resistor in the figure below?

• A. 0.1 A
• B. 0.2 A
• C. 0.3 A
• D. 0.4 A

Answer 1

The correct option is B 0.2 A

As the current flows from a higher potential to a lower potential, the current direction is as shown below from node 1 to node 0.


If voltage at node 0 is $V_0$ then applying Kirchoff's loop law, the current $i$ through the $10\Omega$ is given by:

$i=\frac{10-V_0}{10}$ $⟹$ $10-V_0=10i$ -- (1)


Consider the currents flowing through $20\Omega$ and $30\Omega$ be $i_2$ and $i_3$ respectively. Applying the law similarly for nodes 2 and 3 as shown above, we will get
$V_0-7=20i_1$ -- (2)
$V_0-5=30i_2$ -- (3)
Adding equations (1), (2) we will get $3=10i+20i_1=30i_1+10i_2$ (since $i=i_1+i_2$)
$30i_1+10i_2=3$
$10i_1+40i_2=5$
Solving,
$i=\frac{19}{110}A\approx 0.2A$