Question

In the given circuit diagram calculate the current $I_2$ through the resistor of $10\Omega$.

Answer 1

Between point B and C the resistances are in parallel combination. So voltage difference remains the same .

$V=IR$

$I_1\times 5=I_2\times 10$

$I_1=2I_2$

and equivalent resistance,$R=\frac{5\times 10}{5+10}=\frac{10}{3}$

Current I =$\frac{12}{R}=\frac{12\times 3}{10}=3.6A$

$I=I_1+I_2$

$I=3I_2$

Current through $10\Omega$ is $I_2=\frac{I}{3}=\frac{3.6}{3}=1.2A$