Exam

AIIMS

Year

2018

Date

27 May

Subject

Physics

Chapter

Current Electricity

AIIMS 2018 · 27 May · Physics · Current Electricity

In the circuit shown, the current through the 5 resistor is

In the circuit shown, the current through the $5 \Omega$ resistor is

Step-by-step solution

Applying Kirchhoff's second law for closed loop $A E F B A$, we get $-\left(I_1+I_2\right) \times 5-I_1 \times 2+2=0$
Again, applying Kirchhoff's second law for a closed loop $D E F C D$, we get $-\left(I_1+I_2\right) \times 5-I_2 \times 2+2=0$
On solving, we get $I_1=\frac{1}{6} \mathrm{~A}$ and $I_2=\frac{1}{6} \mathrm{~A}$ $I=I_1+I_2=\frac{1}{6}+\frac{1}{6}=\frac{1}{3} \mathrm{~A}$