Electricity
Electricity is the A-level Physics topic covering current, potential difference, resistance, resistivity, EMF and internal resistance, and the current-voltage characteristics of components such as filament lamps and diodes. It includes two required practicals: finding the resistivity of a wire, and investigating I-V characteristics.
Before you start
Make sure you're comfortable with these topics first:
Method
- Recall the definition of current (I = Q/t, the rate of flow of charge) and potential difference (energy transferred per unit charge, V = W/Q).
- Use I = n x A x v x q to relate current to the number density of charge carriers, cross-sectional area, drift velocity and charge.
- Apply R = resistivity x L / A to calculate the resistance of a wire, converting all lengths and areas to metres and square metres first.
- Use P = I x V, P = I^2 x R and P = V^2/R interchangeably to calculate electrical power, choosing whichever form uses the quantities given.
- For circuits with internal resistance, use EMF = I x (R + r), where R is the external resistance and r is the internal resistance, to include the potential difference lost inside the battery.
- Sketch and interpret I-V characteristic graphs: a straight line through the origin for a resistor at constant temperature, a curve of increasing gradient for a filament lamp, and negligible current below the threshold voltage for a diode.
Worked example
A wire has a resistance of 6.0 ohm and carries a current of 2.5 A. Calculate the power dissipated in the wire, and the potential difference across it.
- Identify the known quantities: I = 2.5 A, R = 6.0 ohm.
- Use P = I^2 x R to calculate the power dissipated.
- Substitute: P = (2.5)^2 x 6.0 = 6.25 x 6.0.
- Evaluate: P = 37.5 W.
- Use V = I x R to calculate the potential difference: V = 2.5 x 6.0.
- Final answer: P = 37.5 W, V = 15 V.
Practice questions
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Q1State the equation linking charge Q, current I and time t.Show answer
Answer: Q = I x t
Q2A current of 3.0 A flows for 40 s. Calculate the charge that flows.Show answer
Answer: 120 C (Q = I x t = 3.0 x 40)
Q3A component has a resistance of 12 ohm when a potential difference of 6.0 V is applied. Calculate the current.Show answer
Answer: 0.50 A (I = V/R = 6.0/12)
Q4Calculate the resistance of a nichrome wire of length 2.0 m and cross-sectional area 4.0 x 10^-7 m^2, given the resistivity of nichrome is 1.1 x 10^-6 ohm metre.Show answer
Answer: 5.5 ohm (R = resistivity x L / A = 1.1 x 10^-6 x 2.0 / 4.0 x 10^-7)
Q5A battery of EMF 9.0 V and internal resistance 1.0 ohm is connected to an external resistor of 8.0 ohm. Calculate the current in the circuit.Show answer
Answer: 1.0 A (I = EMF / (R + r) = 9.0 / (8.0 + 1.0))
Q6Two resistors of 6.0 ohm and 3.0 ohm are connected in parallel. Calculate their combined resistance.Show answer
Answer: 2.0 ohm (1/R = 1/6.0 + 1/3.0 = 3/6.0, so R = 2.0 ohm)
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
A 2.0 kW electric heater is connected to the UK 230 V mains supply. Calculate the current drawn by the heater.
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A student investigates the resistivity of a metal wire of diameter 0.46 mm. The wire has a length of 1.20 m and a resistance of 3.8 ohm. Calculate the resistivity of the wire.
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A student sets up a circuit to find the internal resistance of a battery, using a variable resistor, an ammeter and a voltmeter. The results are: at I = 0.20 A, V = 5.60 V; at I = 0.80 A, V = 4.40 V. Using EMF = V + I x r, calculate the EMF and internal resistance of the battery from these two readings.
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See real A Level Science past-paper questions, with official mark schemes →
Free printable worksheet
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