# Kirchhoff's laws and their applications

Kirchhoff's laws for electrical circuits are (i) The Junction Law and (ii) The Loop Law.

## Kirchhoff's junction law

According to Kirchhoff's junction law, the sum of currents arriving at a node in a circuit is equal to the sum of currents leaving that node. In othe wrods, the algebraic sum of all the currents directed towards a node is zero i.e., $\Sigma_\text{node}\, I_i=0$.

Kirchhoff's junction law is based on charge conservation. In steady state, the charge cannot accumulate.

## Kirchhoff's loop law

According to Kirchhoff's loop law, the algebraic sum of emf in a loop is equal to the algebraic sum of the products of current and resistance. In other words, the algebraic sum of all the potential differences along a closed loop in a circuit is zero i.e., $\Sigma_\text{loop} \Delta\, V_i=0$.

Kirchhoff's loop law is based on energy conservation.

## Battery with internal resistance

Consider a battery of emf $\mathcal{E}$ and internal resistance $r$. When a resistance $R$ is connected across the battery terminals, a current $I$ flows in the circuit. Apply Kirchhoff's loop law to get
\begin{align}
\mathcal{E}=Ir+IR
\end{align}
The potential difference across the resistance $R$ is
\begin{align}
V&=RI=R\frac{\mathcal{E}}{R+r} \\
&=\mathcal{E}\left(1-\frac{r}{R+r}\right)
\end{align}

The potential difference between the terminals of a battery is always less than its emf when it is supplying a current. If, however, no current is being supplied by the battery (i.e., it is an open circut), the potential difference between its terminal is equal to $\mathcal{E}$.

## 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}
& i_1R_1+(i_1-i_2)R_2=V_1,\\
& 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.

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