Q.A battery of emf 4 volt and internal resistance 1 \Omega is connected in parallel with another battery of emf 1 V and internal resistance 1 \Omega (with their like poles connected together). The combination is used to send current through an external resistance of 2 \Omega. Calculate the current through the external resistance.

Kirchhoff's Laws and Circuit Analysis

Complex circuits that cannot be reduced to simple series–parallel combinations are solved with Kirchhoff's two rules:

• Junction (current) rule — the algebraic sum of currents meeting at a node is zero, $\sum I = 0$. It expresses conservation of charge.
• Loop (voltage) rule — around any closed loop the algebraic sum of EMFs and $IR$ drops is zero, $\sum \varepsilon = \sum IR$. It expresses conservation of energy.

To apply them, assign a current to each branch, write one junction equation per independent node and one loop equation per independent mesh, and solve the simultaneous equations; a negative answer simply means the assumed direction was reversed. Where node potentials are given, the current in a branch follows directly from Ohm's law, $I=(V_1-V_2)/R$, and the junction rule fixes an unknown branch current from the others.

These laws handle multi‑cell networks (cells aiding or opposing), unbalanced bridges, and questions such as 'the current through a resistor is unchanged when an extra cell is added', which reduce to setting up and solving the loop and junction equations.

Kirchhoff's junction and loop rules are a core, high-weightage part of the CBSE Class 12 Physics Current Electricity chapter, frequently searched as "Kirchhoff's laws class 12 physics important questions" or "Kirchhoff's current and voltage law examples". Solving multi-loop and multi-cell networks with these rules is one of the most heavily tested numerical skills in board exams as well as JEE Main and NEET.