Recall and use in calculations, the fact that: (a) the sum of the currents entering a junction in a parallel circuit is equal to the sum of the currents that leave the junction (b) the total p.d. across the components in a series circuit is equal to

4.3.2 Series and Parallel Circuits

Learning Objectives

• Recall and use the fact that the sum of the currents entering a junction in a parallel circuit is equal to the sum of the currents that leave the junction.
• Recall and use the fact that the total p.d. across the components in a series circuit is equal to the sum of the individual p.d.s across each component.
• Recall and use the fact that the p.d. across an arrangement of parallel resistances is the same as the p.d. across any one branch of that arrangement.

1. Junction (Current) Rule – Parallel Circuits

At any junction (node) the algebraic sum of currents is zero. For a simple parallel network:

\$\sum I{\text{in}} = \sum I{\text{out}}\$

In practice, if a current \$I\$ enters a junction and splits into \$I1\$ and \$I2\$ in two branches, then

\$I = I1 + I2\$

2. Series \cdot oltage Rule

When resistors are connected end‑to‑end, the same current flows through each, and the total potential difference (p.d.) across the series combination equals the sum of the p.d.s across each resistor.

\$V{\text{total}} = V1 + V2 + V3 + \dots\$

Using Ohm’s law \$V = IR\$, the series rule can also be written as

\$I R{\text{total}} = I R1 + I R2 + I R3 + \dots\$

which leads to the familiar result

\$R{\text{total}} = R1 + R2 + R3 + \dots\$

3. Parallel \cdot oltage Rule

All branches of a parallel network experience the same potential difference.

\$V{\text{across\,parallel}} = V{\text{branch\,1}} = V_{\text{branch\,2}} = \dots\$

Consequently the equivalent resistance of parallel resistors is given by

\$\frac{1}{R{\text{total}}}= \frac{1}{R1}+ \frac{1}{R2}+ \frac{1}{R3}+ \dots\$

4. Example Calculations

Example 1 – Parallel Circuit Current Check

Three resistors are connected in parallel across a 12 V battery:

• \$R_1 = 4\;\Omega\$
• \$R_2 = 6\;\Omega\$
• \$R_3 = 12\;\Omega\$

Calculate the individual branch currents and verify the junction rule.

1. Find each branch current using \$I = V/R\$:

   • \$I_1 = \dfrac{12}{4}=3.0\;\text{A}\$
   • \$I_2 = \dfrac{12}{6}=2.0\;\text{A}\$
   • \$I_3 = \dfrac{12}{12}=1.0\;\text{A}\$

2. Sum of branch currents:

   \$I{\text{total}} = I1 + I2 + I3 = 3.0 + 2.0 + 1.0 = 6.0\;\text{A}\$

3. Since the battery supplies \$6.0\;\text{A}\$ to the junction, the total current leaving the junction is also \$6.0\;\text{A}\$ – the junction rule is satisfied.

Example 2 – Series Circuit \cdot oltage Check

A series chain of three resistors carries a current of \$0.5\;\text{A}\$:

• \$R_1 = 10\;\Omega\$
• \$R_2 = 20\;\Omega\$
• \$R_3 = 30\;\Omega\$

Find the total p.d. supplied and verify that it equals the sum of the individual p.d.s.

1. Individual voltage drops:

   • \$V1 = I R1 = 0.5 \times 10 = 5.0\;\text{V}\$
   • \$V2 = I R2 = 0.5 \times 20 = 10.0\;\text{V}\$
   • \$V3 = I R3 = 0.5 \times 30 = 15.0\;\text{V}\$

2. Sum of voltage drops:

   \$V{\text{total}} = V1 + V2 + V3 = 5.0 + 10.0 + 15.0 = 30.0\;\text{V}\$

3. Equivalent resistance \$R{\text{total}} = 10 + 20 + 30 = 60\;\Omega\$, so \$V{\text{total}} = I R_{\text{total}} = 0.5 \times 60 = 30.0\;\text{V}\$ – confirming the series voltage rule.

5. Summary Table of Key Relations