Example 4: Observe the given circuit and calculate the resultant resistance.
Solution: From the given circuit
R₁ =4Ω, R₂ =12Ω,are connected in parallel their resultant resistance is
R=R₁ R₂ / R₁ +R₂=4×12/4+12 = 48/16 = 3Ω
R1, R2 are connected to R3 in series then resultaint resistance is R = 3 + 7 = 10 Ω

Though the methods discussed in the previous section for replacing series and parallel combination of resistors by their equivalent resistances are very useful for simplifying many combinations of resistors, they are not sufficient for the analysis of many simple circuits particularly those containing more than one battery.

Let us see.

Kirchhoff’s laws

Two simple rules called Kirchhoff’s rules are applicable to any DC circuit containing batteries and resistors connected in any way.

Junction Law

see figure 19.
We have seen that the current divides at P. The current drawn from the battery is equal to the sum of the currents through the resistors. P is called junction. The junction is a point where three or more conducting wires meet.

At any junction point in a circuit where the current can divide, the sum of the currents into the junction must equal the sum of the currents leaving the junction. This means that there is no accumulation of electric charges at any junction in a circuit.

From figure 23, we have

I₁+I₄+I₆ = I₅+I₂+I₃

This law is based on the conservation of charge.

Loop Law

The algebraic sum of the increases and decreases in potential difference across various components of the circuit in a closed circuit loop must be