Question

Find the current in $6\Omega$ resistor in the circuit shown in the figure.

• A. $1\text{A}$
• B. $0.5\text{A}$
• C. $2\text{A}$
• D. $2.5\text{A}$

Answer 1

The correct option is B $0.5\text{A}$

Suppose the currents in the loops are as shown in the figure.


Applying KVL in loop $1$ in clockwise direction starting from negative terminal of $4.5\text{V}$ battery.

$+4.5-3I_1-6(I_1-I_2)=0$

$\Rightarrow 9I_1-6I_2=4.5...\left(1\right)$

Similarly, for loop $2$,

$-10I_2-3+6(I_1-I_2)=0$

$\Rightarrow$ $6I_1-16I_2=3...\left(2\right)$

On solving equations $\left(1\right)$ and $\left(2\right)$, we get,

$\Rightarrow I_1=0.5\text{A}$ and $I_2=0\text{A}$

Therefore, the current through the $6\Omega$ resistor is $I_1-I_2=0.5-0=0.5\text{A}$

Hence, option $\left(B\right)$ is the correct answer.