Students should be able to apply the rules for current, potential difference and effective resistance in series and parallel circuits, and use \(R = \dfrac{V}{I}\) to solve whole-circuit problems.
Components can be connected in series, in parallel, or in a combination of both. The rules for current and potential difference depend on the arrangement of the components.
In a series circuit, components are connected one after another in a single path. There are no branches, so all charges must pass through every component.
The current is the same at every point in a series circuit:
\[I_{\text{total}} = I_1 = I_2 = I_3\]
This is because there is only one path for charge to flow. Charge does not get used up by a component.
The sum of the potential differences across the components is equal to the potential difference across the whole circuit:
\[V_{\text{supply}} = V_1 + V_2 + V_3\]
The supply transfers energy to the charges. The charges then transfer this energy to the components as they pass through them.
The effective resistance of resistors in series is the sum of their individual resistances:
\[R_{\text{total}} = R_1 + R_2 + R_3\]
Adding more resistors in series increases the total resistance and decreases the current for the same supply potential difference.
In a parallel circuit, components are connected on separate branches. Each branch provides a different path for charge to flow.
The current from the source is equal to the sum of the currents in the separate branches:
\[I_{\text{total}} = I_1 + I_2 + I_3\]
At a junction, current splits between the branches. When the branches meet again, the branch currents recombine.
The potential difference across each parallel branch is the same as the potential difference across the supply:
\[V_{\text{supply}} = V_1 = V_2 = V_3\]
This is because each branch is connected across the same two points of the circuit.
For resistors in parallel, the reciprocal of the effective resistance is the sum of the reciprocals of the individual resistances:
\[\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]
For two resistors in parallel, this can also be written as:
\[R_{\text{total}} = \frac{R_1R_2}{R_1 + R_2}\]
The effective resistance of resistors in parallel is always less than the smallest individual resistance. Adding another parallel branch gives charge an extra path, so the total current from the supply increases.
The relationship between resistance, potential difference and current is:
\[R = \frac{V}{I}\]
It can be rearranged as \(V = IR\) and \(I = \dfrac{V}{R}\). In circuit problems, first decide whether components are in series or parallel. Then apply the correct current and potential difference rules before using \(R = \dfrac{V}{I}\).
A 6.0 V battery is connected to two resistors, \(2.0\,\Omega\) and \(4.0\,\Omega\), in series.
The total resistance is:
\[R_{\text{total}} = 2.0 + 4.0 = 6.0\,\Omega\]
The current in the circuit is:
\[I = \frac{V}{R} = \frac{6.0}{6.0} = 1.0\,\text{A}\]
The same current flows through both resistors. The potential differences are \(V_1 = 1.0 \times 2.0 = 2.0\,\text{V}\) and \(V_2 = 1.0 \times 4.0 = 4.0\,\text{V}\). Their sum is \(6.0\,\text{V}\), the battery voltage.
A 12 V supply is connected across two resistors, \(6.0\,\Omega\) and \(3.0\,\Omega\), in parallel.
Each resistor has the full 12 V across it. The branch currents are:
\[I_1 = \frac{12}{6.0} = 2.0\,\text{A}\]
\[I_2 = \frac{12}{3.0} = 4.0\,\text{A}\]
The current from the source is \(I_{\text{total}} = 2.0 + 4.0 = 6.0\,\text{A}\). The effective resistance is therefore:
\[R_{\text{total}} = \frac{12}{6.0} = 2.0\,\Omega\]
| Quantity | Series circuit | Parallel circuit |
|---|---|---|
| Current | Same through every component. | Splits between branches; branch currents add to the source current. |
| Potential difference | Shared between components; individual p.d.s add to the supply p.d. | Same across every branch. |
| Effective resistance | \(R_{\text{total}} = R_1 + R_2 + \cdots\) | \(\dfrac{1}{R_{\text{total}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots\) |