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Series and Parallel Circuits: Equations & Rules

Circuit Basics: Current, Potential Difference, & Resistance

Before diving into rules, you must command the definitions of the three key quantities of circuit calculations:

  • Current ($I$, measured in Amperes, $A$): The rate of flow of electrical charge.
  • Potential Difference ($V$, measured in Volts, $V$): The work done per unit charge (often called voltage).
  • Resistance ($R$, measured in Ohms, $\Omega$): The measure of how hard it is for current to flow.

Ohm's Law links these three elements together in a crucial equation:

$$V = I \times R$$

Rules of Series Circuits

A series circuit has only **one path** for current to flow. Think of it as a single loop. Here are the three non-negotiable rules for series calculations:

1

Current is the SAME everywhere

$$I_{total} = I_1 = I_2 = I_3$$

2

Potential Difference splits across components

$$V_{total} = V_1 + V_2 + V_3$$

3

Total Resistance is the sum of all resistors

$$R_{total} = R_1 + R_2 + R_3$$

Rules of Parallel Circuits

A parallel circuit contains **branches** or multiple loops. The current splits, but the potential difference remains constant across each parallel branch:

1

Potential Difference is the SAME across each branch

$$V_{total} = V_{branch 1} = V_{branch 2}$$

2

Current splits across branches

$$I_{total} = I_{branch 1} + I_{branch 2}$$

Crucial Fact: The total resistance of two or more resistors in parallel is **always less** than the resistance of the smallest individual resistor. This is because adding branches provides more pathways for charges to flow.

Worked Exam Example

Worked GCSE Example

Question: A $12\text{V}$ battery is connected in series with a $4\Omega$ resistor and a $2\Omega$ resistor.
1. Calculate the total resistance of the circuit.
2. Calculate the current flowing through the $4\Omega$ resistor.

Solution:

  1. Since the components are connected in series, we add their resistances: $$R_{total} = R_1 + R_2 = 4\Omega + 2\Omega = 6\Omega$$
  2. Use Ohm's Law ($V = I \times R$) to find the total current: $$I = \frac{V}{R_{total}} = \frac{12\text{V}}{6\Omega} = 2\text{A}$$
  3. Since it is a series circuit, the current is the same everywhere. Therefore, the current through the $4\Omega$ resistor is also $2\text{A}$.

Answers: Total Resistance = $6\Omega$, Current = $2\text{A}$.

Frequently Asked Questions

Adding resistors in parallel adds more loops/branches for the electric charge to flow through. Think of it like adding extra lanes to a highway; even if the new lanes have speed limits, the overall capacity increases, allowing more cars (current) to pass through, which corresponds to a lower overall resistance.

In a series circuit, if one bulb breaks, the entire circuit is broken and all bulbs go out. In a parallel circuit, each branch is independent; if a bulb on one branch breaks, bulbs on other branches will continue to shine because their loops are still complete.

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