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.

Again applying kirchhoff's loop law for node 2 and 3 respectively.

$\frac{V_0-7}{20}=i_2$
or,

$V_0-7=20i_2$ ------ (2)

$\frac{V_0-5}{30}=i_3$
or,

$V_0-5=30i_3$ ------ (3)

Adding equations (1) & (2)
we will get
$3=10i+20i_2$
$3=30i_2+10i_3$ (as $i=i_2+i_3$)
$30i_2+10i_3=3$ --------- (4)

Now adding equation (1) and (3), we will get
$5=10i+30i_3$
$5=10i_2+40i_3$ (as $i=i_2+i_3$)
$10i_2+40i_3=5$ --------- (5)

Solving equation (4) and (5), we get
$i_2=\frac{7}{110}$
$i_3=\frac{12}{110}$
and,
$i=i_2+i_3=\frac{19}{110}A\approx 0.2A$