Series and parallel circuits: current, voltage and resistance

A 9 V battery is connected to a 6 Ω and a 3 Ω resistor in series.

1. Total resistance: $6 + 3 = 9$ Ω.
2. Current: $I = 9 \div 9 = 1$ A, the same through both resistors.
3. Potential difference across each: $V_1 = 1 \times 6 = 6$ V and $V_2 = 1 \times 3 = 3$ V.

Check: $6 + 3 = 9$ V, the battery's potential difference. The bigger resistor takes the bigger share.

The same resistors are connected in parallel to a 6 V supply.

1. Each branch has the full 6 V across it.
2. Currents: $I_1 = 6 \div 6 = 1$ A and $I_2 = 6 \div 3 = 2$ A.
3. Total current: $1 + 2 = 3$ A, so the total resistance is $6 \div 3 = 2$ Ω.

The formula $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ gives the same answer: $\dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}$, so $R = 2$ Ω. (This formula is in Cambridge IGCSE Extended; GCSE students explain why the total is less than the smallest resistor.)

Adding a branch gives the current another path, so more current flows from the same supply. More current for the same potential difference means less total resistance. This is why plugging in more appliances at home draws more current from the supply.

• An ammeter goes in series, in the part of the circuit where you want the current.
• A voltmeter goes in parallel across the component.

Ammeters have very low resistance and voltmeters very high resistance, so they don't change what they're measuring much.

• Adding resistances in parallel as if they were in series.
• Saying current is "used up" in a series circuit: it's the same all the way round.
• Forgetting that each parallel branch gets the full supply potential difference.

Check your numbers with the Ohm's law calculator.

1. A 4 Ω and an 8 Ω resistor are in series with a 6 V battery. Find the current and the potential difference across each.
2. Two 4 Ω resistors are in parallel with a 6 V supply. Find the current in each branch and the total resistance.

Answers: 12 Ω total, 0.5 A, 2 V and 4 V; 1.5 A in each branch, 3 A in total, 2 Ω.