Series & Parallel Circuits
Series Circuits
In a series circuit, all components are connected end to end in a single loop. There is only one path for the current to follow.
In a series circuit:
Current is the same through every component. There is only one path, so the same charge flows through each part: \(I_{\text{total}} = I_1 = I_2 = I_3\).
Potential difference is shared between the components. The total potential difference of the supply equals the sum of the potential differences across each component: \(V_{\text{total}} = V_1 + V_2 + V_3\).
Resistance adds up. The total resistance is the sum of the individual resistances: \(R_{\text{total}} = R_1 + R_2 + R_3\).
In a series circuit: current is the same everywhere, potential differences add up to the supply voltage, and resistances add up.
Two resistors of 10 \(\Omega\) and 20 \(\Omega\) are connected in series to a 6 V battery. Calculate the total resistance and the current in the circuit.
Total resistance: \(R_{\text{total}} = 10 + 20 = 30 \; \Omega\)
Current: \(I = \frac{V}{R} = \frac{6}{30} = 0.2\) A
Parallel Circuits
In a parallel circuit, components are connected across each other on separate branches. The current has more than one path to follow.
In a parallel circuit:
Current splits at junctions. The total current from the supply equals the sum of the currents through each branch: \(I_{\text{total}} = I_1 + I_2 + I_3\). Current is conserved at every junction.
Potential difference is the same across each branch. Every component connected in parallel has the same potential difference across it: \(V_{\text{total}} = V_1 = V_2 = V_3\).
Resistance decreases. Adding more branches gives the current more paths to flow through, so the total resistance is always less than the smallest individual resistance.
In a parallel circuit: the potential difference is the same across each branch, currents through each branch add up to the total current, and the total resistance is less than the smallest individual resistance.
Two resistors are connected in parallel to a 12 V battery. One has a resistance of 6 \(\Omega\) and the other 12 \(\Omega\). Calculate the current through each resistor and the total current.
Both resistors have 12 V across them (parallel circuit rule).
Current through 6 \(\Omega\) resistor: \(I_1 = \frac{V}{R} = \frac{12}{6} = 2\) A
Current through 12 \(\Omega\) resistor: \(I_2 = \frac{V}{R} = \frac{12}{12} = 1\) A
Total current: \(I_{\text{total}} = 2 + 1 = 3\) A
Energy in Circuits
When charge flows through a resistor, the electrical energy is transferred to the thermal store of the resistor, which heats up. This happens because the moving electrons collide with the metal ions in the lattice, transferring energy to them and making them vibrate more. This energy is then dissipated to the thermal store of the surroundings.
We can reduce unwanted heating by using wires with lower resistance. Thicker wires and shorter wires have lower resistance, which means less energy is wasted as heat.
Electrical Power
Power is the rate at which energy is transferred, measured in watts (W). One watt equals one joule of energy transferred per second. There are three key power equations:
\(P = \frac{E}{t}\) — power equals energy transferred divided by time
\(P = I \times V\) — power equals current times potential difference
\(P = I^2 \times R\) — power equals current squared times resistance
A heater has a current of 3 A and a resistance of 40 \(\Omega\). Calculate the power of the heater.
\(P = I^2 \times R = 3^2 \times 40 = 9 \times 40 = 360\) W
The three power equations are: \(P = \frac{E}{t}\), \(P = I \times V\), and \(P = I^2 \times R\). Choose the equation that uses the values you are given in the question.
Exam Tip: Learn the series and parallel rules as a set of contrasts. In series, current stays the same and voltages add up. In parallel, voltage stays the same and currents add up. You need to be able to apply these rules to unfamiliar circuits.
Series and parallel circuits follow clear, predictable rules. Knowing how current, potential difference, and resistance behave in each type of circuit — and being able to calculate power and energy transfer — is essential for the electricity section of the exam.