Ch 2 · 10 D.C. Circuits · A Level Physics 9702
Topic 2 of 2 in Electricity — A Level Physics 9702 — 89 lessons.
Introduction
Real batteries get warm, their voltage drops when you draw a big current, and complicated networks of resistors still obey two simple conservation laws. This topic gives you the tools to analyse any d.c. circuit. (Syllabus topic 10: sections 10.1–10.3.)
Key ideas — 10.1 (a) Circuit symbols and diagrams
You must recall and use the circuit symbols listed in the syllabus (cell, battery, power supply, switch, lamp, fixed and variable resistor, thermistor, LDR, potentiometer, diode, LED, ammeter, voltmeter, galvanometer, fuse, earth, and others) and draw and interpret circuits built from them. Draw wires with a ruler; show junctions with a dot.
🧑🏫 Teacher's explanation
Think of a circuit like a water plumbing system. The battery is the pump that adds pressure (energy) to the water, the wires are the pipes, and the resistors are narrow sections of pipe that slow the flow down. To understand a circuit, you need to be able to read the 'blueprint'—the circuit diagram—to see how the water (charge) travels and where it might get stuck or diverted.
📖 Definitions
🔍 Worked example — Easy
Identify the number of components in a circuit containing a single cell, a closed switch, a lamp, and an ammeter connected in series.
🔍 Worked example — Standard
A student draws a circuit diagram. They connect a battery to a resistor and a voltmeter in parallel. How many junctions (dots) must be drawn to show the connection between the voltmeter and the resistor?
🔍 Worked example — Exam-style
A circuit diagram shows a battery connected to a variable resistor and a fixed resistor in series. If the diagram shows the wires crossing without a dot, what is the physical implication for the circuit?
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 10.1 (b) E.m.f. and p.d.
The e.m.f. E of a source is the energy transferred per unit charge (from other forms, e.g. chemical, to electrical) in driving charge round a complete circuit.
🧑🏫 Teacher's explanation
Imagine you are at a water park. The pump that pushes water from the bottom pool up to the top of the slide represents the electromotive force (e.m.f.). It provides the 'push' or the total energy required to get the water moving through the whole system. The water's height at the top is like the total energy available.
📖 Definitions
🔍 Worked example — Easy
A battery has an e.m.f. of 12.0 V. When a resistor is connected, the terminal p.d. is measured to be 11.2 V. Calculate the 'lost volts' (the p.d. across the internal resistance).
🔍 Worked example — Standard
A cell has an e.m.f. of 1.5 V and an internal resistance of 0.5Omega. Calculate the terminal p.d. when a current of 0.4 A flows through the circuit.
🔍 Worked example — Exam-style
A battery with e.m.f. E and internal resistance r is connected to an external resistor R. Derive an expression for the terminal p.d. V in terms of E, r, and R.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
(a) Calculate the current in the circuit. [2] (b) Calculate the terminal potential difference across the 4.8Omega resistor. [2] [4 marks]
Key ideas — 10.1 (c) Internal resistance
A real source has internal resistance r: some energy per unit charge is dissipated inside the source itself. With an external resistance R and current I:
🧑🏫 Teacher's explanation
Imagine you are using a water pump to move water through a garden hose. The pump provides a certain amount of pressure (the e.m.f.), but the pump itself has some friction and internal narrowings that resist the flow. This internal friction is like the 'internal resistance'. Because some energy is 'wasted' just getting the water through the pump, the pressure available at the end of the hose (the terminal p.d.) is always slightly less than what the pump originally provided.
📖 Definitions
🔍 Worked example — Easy
A battery has an e.m.f. of 1.5 V and an internal resistance of 0.5Omega. Calculate the terminal potential difference when a 2.0Omega resistor is connected to the battery.
🔍 Worked example — Standard
A cell with e.m.f. mathcalE and internal resistance r is connected to a variable resistor R. When R = 5.0Omega, the terminal p.d. is 4.0 V and the current is 0.50 A. Calculate the e.m.f. and the internal resistance.
🔍 Worked example — Exam-style
A graph of terminal potential difference V against current I for a battery is plotted. The line passes through (0.0, 12.0) and (2.0, 10.0). Determine the e.m.f. and the internal resistance of the battery.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 10.2 (a) Kirchhoff's laws
Kirchhoff's first law: the sum of the currents entering a junction equals the sum of the currents leaving it. This is a consequence of conservation of charge (charge cannot build up at a junction).
🧑🏫 Teacher's explanation
Think of electric current like water flowing through pipes. If you have a T-junction in a pipe, the amount of water flowing into that junction per second must exactly equal the amount flowing out. If it didn't, water would be piling up inside the junction, which is impossible in a steady state. This is Kirchhoff's First Law, and it is fundamentally about the conservation of charge.
📖 Definitions
🔍 Worked example — Easy
A circuit junction has three wires. A current of 0.50 A enters the junction. Two wires leave the junction; one carries 0.25 A and the other carries I. Calculate the value of I.
🔍 Worked example — Standard
A closed loop contains a battery with e.m.f. 12.0 V and two resistors with potential differences V_1 = 4.5 V and V_2 across them. Calculate V_2.
🔍 Worked example — Exam-style
A circuit consists of a 15.0 V e.m.f. source connected to two resistors in series, R_1 = 10.0Omega and R_2 = 20.0Omega. Calculate the current I flowing through the circuit using Kirchhoff's laws.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) State Kirchhoff's Second Law for the closed loop containing the battery and the resistor. [1] (b) Show that the terminal potential difference V across the resistor is given by V = E - Ir, where I is the current in the circuit. [2] [3 marks]
Key ideas — 10.2 (b) Deriving the series and parallel formulae
Series: the same current I flows through each resistor. By the second law, the total p.d. V = V_1 + V_2 + … = IR_1 + IR_2 + …. Since V = IR_total:
🧑🏫 Teacher's explanation
Imagine water flowing through a pipe system. In a series circuit, the water flows through one pipe, then another, then another, all in a single line. The total resistance is just the sum of every obstacle the water hits. Because there is only one path, the flow rate (current) must be the same everywhere.
📖 Definitions
🔍 Worked example — Easy
Two resistors, R_1 = 4.0 Omega and R_2 = 6.0 Omega, are connected in series to a battery. Calculate the total resistance R_total.
🔍 Worked example — Standard
Two resistors, R_1 = 12 Omega and R_2 = 4 Omega, are connected in parallel. Calculate the total resistance R_total.
🔍 Worked example — Exam-style
A circuit consists of a 12.0 V battery connected to three resistors in parallel: R_1 = 10.0 Omega, R_2 = 20.0 Omega, and R_3 = 30.0 Omega. Calculate the total current I_total flowing from the battery.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — 10.2 (c) Solving circuits
🧑🏫 Teacher's explanation
Think of a circuit like a water plumbing system. Kirchhoff's First Law is like saying: if 5 liters of water enter a junction in a pipe, 5 liters must exit through the other pipes. Charge is never lost; it just splits up. Kirchhoff's Second Law is like saying: if you start at a water tank, go through a series of pumps and pipes, and return to the same tank, the total energy you gained from the pumps must equal the total energy lost to friction in the pipes. In electrical terms, the energy supplied by the battery (e.m.f.) equals the energy used by the resistors.
📖 Definitions
🔍 Worked example — Easy
A simple series circuit consists of a 12.0 V battery and two resistors, R_1 = 4.0Omega and R_2 = 8.0Omega. Calculate the current in the circuit.
🔍 Worked example — Standard
A circuit has a 24.0 V battery and two resistors in parallel, R_1 = 12.0Omega and R_2 = 6.0Omega. Calculate the total current leaving the battery.
🔍 Worked example — Exam-style
A circuit contains a battery with e.m.f. varepsilon = 10.0 V and internal resistance r = 1.0Omega. It is connected to an external resistor R = 4.0Omega. Calculate the terminal potential difference V.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 10.3 (a) Potential dividers
Two resistors in series across a p.d. V_in share it in proportion to their resistances:
🧑🏫 Teacher's explanation
Imagine a water pipe system where a large water pressure is split between two different valves. If one valve is much harder to open (higher resistance) than the other, most of the pressure drop will occur across that harder valve. In a DC circuit, voltage (potential difference) behaves just like that pressure.
📖 Definitions
🔍 Worked example — Easy
A 12.0 V battery is connected to two resistors in series: R_1 = 400 Omega and R_2 = 800 Omega. Calculate the potential difference across R_2.
🔍 Worked example — Standard
A potential divider uses a fixed resistor R_1 = 2.0 k Omega and an LDR as R_2. The input voltage is 6.0 V. When light shines on the LDR, its resistance is 1.0 k Omega. Calculate the output voltage V_out across the LDR.
🔍 Worked example — Exam-style
An LDR is used in a potential divider with a 10.0 V source and a fixed 5.0 k Omega resistor. The output voltage is taken across the LDR. If the LDR resistance increases from 5.0 k Omega to 20.0 k Omega as it gets darker, calculate the change in output voltage Δ V_out.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — 10.3 (b) The potentiometer and null methods
A potentiometer here is a uniform resistance wire (often 1 m long) with a steady current from a driver cell, so the p.d. along it is proportional to length.
🧑🏫 Teacher's explanation
Imagine you are trying to measure the height of a person using a ruler, but the ruler itself is slightly tilted. Instead of reading the numbers on the ruler, you use a second, identical ruler to find the exact point where the person's height matches the ruler's markings. This is the essence of a 'null method'.
📖 Definitions
🔍 Worked example — Easy
A potentiometer wire has a total length of 1.00 m. A cell with e.m.f. E is balanced against a length of 0.40 m of the wire. Calculate the ratio of the cell's e.m.f. to the total p.d. across the entire wire.
🔍 Worked example — Standard
Two cells, E_1 = 1.50 V and E_2 = 3.00 V, are balanced against a potentiometer wire at lengths l_1 = 40.0 cm and l_2 = x cm respectively. Calculate the value of x.
🔍 Worked example — Exam-style
An unknown e.m.f. E_x is balanced at l_x = 0.65 m. A standard cell of E_s = 1.50 V is balanced at l_s = 0.30 m on the same wire. Calculate E_x.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Example 1: internal resistance. A cell of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor.
🔍 Worked example — Easy
A cell has an e.m.f. of 1.5 V and an internal resistance of 0.50 Omega. It is connected to a 2.5 Omega resistor. Calculate the terminal potential difference.
🔍 Worked example — Standard
A 12 V battery is connected to a 4.0 Omega resistor in series with a parallel combination of a 6.0 Omega and a 12 Omega resistor. Calculate the current through the 6.0 Omega resistor.
🔍 Worked example — Exam-style
A potentiometer is used to compare two cells. A standard cell of e.m.f. 1.02 V balances at a length of 51.0 cm. An unknown cell balances at 75.0 cm. Calculate the e.m.f. of the unknown cell.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) E.m.f.: energy transferred per unit charge from other forms to electrical in driving charge round a complete circuit; p.d.: energy transferred per unit charge from electrical to other forms in a component. (2) r = (9.0 - 8.1)/0.60 = 1.5\ Omega. (3) In a closed loop, the sum of the e.m.f.s equals the sum of the p.d.s; conservation of energy. (4) Same p.d. V across each; I = I_1 + I_2 (first law) = V/R_1 + V/R_2 = V/R, so 1/R = 1/R_1 + 1/R_2. (5) At balance no current flows through the cell, so there is no Ir drop across its internal resistance.