use the formula for the combined resistance of two or more resistors in parallel

Kirchhoff’s Laws

Kirchhoff’s laws are fundamental tools for analysing electric circuits. They allow us to relate currents and voltages in complex networks, especially when resistors are connected in series and parallel.

Learning Objective

By the end of this lesson you should be able to use the formula for the combined resistance of two or more resistors in parallel while applying Kirchhoff’s laws.

Kirchhoff’s Current Law (KCL)

KCL states that the algebraic sum of currents meeting at a junction is zero:

\$\sum{k=1}^{n} Ik = 0\$

In practice this means that the total current entering a node equals the total current leaving the node.

Kirchhoff’s \cdot oltage Law (K \cdot L)

K \cdot L states that the algebraic sum of the potential differences (voltages) around any closed loop is zero:

\$\sum{k=1}^{n} Vk = 0\$

This reflects the conservation of energy for charges moving around a loop.

Applying KCL and K \cdot L to Parallel Resistors

When several resistors are connected in parallel, they share the same voltage across their terminals but carry different currents. Using KCL at the junction where the parallel branch splits and K \cdot L around each branch, we can derive the combined (equivalent) resistance.

Derivation of the Parallel‑Resistance Formula

1. Consider \$n\$ resistors \$R1, R2, \dots , R_n\$ connected in parallel across a voltage source \$V\$.
2. By Ohm’s law, the current through each resistor is \$Ik = \dfrac{V}{Rk}\$.
3. The total current supplied by the source is the sum of the branch currents (KCL):

   \$I{\text{total}} = \sum{k=1}^{n} Ik = \sum{k=1}^{n} \frac{V}{R_k}\$

4. Define the equivalent resistance \$R{\text{eq}}\$ such that \$I{\text{total}} = \dfrac{V}{R_{\text{eq}}}\$.
5. Equating the two expressions for \$I_{\text{total}}\$ and cancelling \$V\$ gives:

   \$\frac{1}{R{\text{eq}}} = \sum{k=1}^{n} \frac{1}{R_k}\$

Formula Summary

Worked Example

Find the equivalent resistance of a circuit where \$R1 = 4\ \Omega\$, \$R2 = 6\ \Omega\$, and \$R_3 = 12\ \Omega\$ are connected in parallel.

1. Write the reciprocal sum:

   \$\frac{1}{R_{\text{eq}}}= \frac{1}{4} + \frac{1}{6} + \frac{1}{12}\$

2. Calculate each term:

   \$\frac{1}{4}=0.250,\quad \frac{1}{6}=0.167,\quad \frac{1}{12}=0.083\$

3. Add them:

   \$\frac{1}{R_{\text{eq}}}=0.250+0.167+0.083 = 0.500\$

4. Invert to obtain \$R_{\text{eq}}\$:

   \$R_{\text{eq}} = \frac{1}{0.500}=2\ \Omega\$

Common Pitfalls

• Adding resistances directly for parallel branches (the opposite of series).
• Forgetting that the voltage across each parallel resistor is the same.
• Neglecting to include all branches when applying KCL at a node.

Summary

Kirchhoff’s laws provide a systematic way to analyse circuits. When resistors are in parallel, KCL ensures the total current is the sum of branch currents, while K \cdot L guarantees a common voltage across each branch. Combining these principles leads to the parallel‑resistance formula, a crucial tool for A‑Level physics.

Practice Questions

1. Three resistors \$5\ \Omega\$, \$10\ \Omega\$ and \$20\ \Omega\$ are connected in parallel. Calculate \$R_{\text{eq}}\$.
2. A \$12\ \text{V}\$ battery supplies a parallel network of \$R1 = 3\ \Omega\$ and \$R2 = 6\ \Omega\$. Determine the total current drawn from the battery.
3. In a circuit, a \$2\ \Omega\$ resistor is in series with a parallel combination of \$4\ \Omega\$ and \$12\ \Omega\$. Find the total resistance of the circuit.