Question

In the circuit shown in figure, the current flowing through $5\Omega$ resistance is

• A. $0.5A$
• B. $0.6A$
• C. $0.9A$
• D. $1.5A$

Answer 1

The correct option is B $0.6A$

$\text{Step 1: Apply Kirchhoff's current law [Refer Fig.]}$

Apply Kirchhoff's current law at junction A,

$2.1A=I_1+I_2$ $....\left(1\right)$

$\text{Step 2: Apply Kirchhoff's Voltage law}$

Apply KVL in the loop (in clockwise direction)

$8I_1+2I_1-5I_2-20I_2=0$

$\Rightarrow 10I_1=25I_2$

$\Rightarrow I_1=2.5I_2$ $....\left(2\right)$

$\text{Step 3: Solving equations}$

Equation $\left(1\right)$ and $\left(2\right)\Rightarrow$$2.1=2.5I_2+I_2$

$\Rightarrow I_2=\frac{2.1}{3.5}=0.6A$

Current through $20\Omega$ and $5\Omega$ will be the same i.e. $0.6A$ as they are connected in series.

Hence, Option $B$ is correct

$\text{Alternate solution:}$

In a parallel combination of resistor $R_1$ and $R_2$, we know that current through $R_2$ is given by-

$I_2=\frac{R_1}{R_1+R_2}I$,

In this case, $R_1=6\Omega +4\Omega =10\Omega$ and $R_2=20\Omega +5\Omega =25\Omega$

$\Rightarrow I_2=\frac{10}{10+25}\times 2.1A=0.6A$

This is current through $5\Omega$ resistor.