Q.Two cells of EMF $8\ \text{V}$ and $10\ \text{V}$ (each branch, including the cell's own internal resistance, has a net resistance of $2\ \Omega$) are connected with their positive terminals joined to a common node $A$. Node $A$ is joined to the common negative node $B$ through a resistor of $2\ \Omega$. Using Kirchhoff's laws, find the current supplied by each cell and the current through the $2\ \Omega$ resistor.

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. …