# Kirchhoff's Laws and Simple Applications

## Problems from IIT JEE

**Problem (IIT JEE 2014):**
Two ideal batteries of *emf* $V_1$ and $V_2$ and three resistances $R_1$, $R_2$ and $R_3$ are connected as shown in the figure. The current in resistance $R_2$ would be zero if,

- $V_1=V_2$ and $R_1=R_2=R_3$
- $V_1=V_2$ and $R_1=2R_2=R_3$
- $V_1=2V_2$ and $2R_1=2R_2=R_3$
- $2V_1=V_2$ and $2R_1=R_2=R_3$

**Solution:**
Let $i_1$ and $i_2$ be the current as shown in figure. Apply Kirchhoff's law in the loop ABCDA and CEFDC to get,
\begin{alignat}{2}
\label{twb:eqn:1}
& i_1R_1+(i_1-i_2)R_2=V_1,\\
\label{twb:eqn:2}
& i_2R_3-(i_1-i_2)R_2=V_2.
\end{alignat}
Multiply first equation by $R_3$ and second equation by $R_1$ and then subtract to get current through $R_2$ as,
\begin{alignat}{2}
(i_1-i_2)=\frac{V_1 R_3-V_2 R_1}{R_1R_3+R_2R_3-R_1R_2}. \nonumber
\end{alignat}
The current through $R_2$ becomes zero when $V_1 R_3=V_2 R_1$. Thus, correct options are A, B, and D.