Kirchhoff's laws and their applications
By 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.
