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CIE 9702 Physics · AS · Topic 10

D.C. circuits

Clear, syllabus-mapped CIE 9702 Physics revision notes on d.c. circuits: explanations, worked examples and exam technique, then a free targeted practice drill.

CIE 9702 PhysicsASFree revision notes
Contents: 7 sections

Syllabus points

Kirchhoff's laws

First law: the sum of the currents into a junction equals the sum of the currents out. This is conservation of charge.

Second law: around any closed loop, the sum of the e.m.f.s equals the sum of the p.d.s. This is conservation of energy.

Naming the conserved quantity is usually worth a mark, and the two are easy to swap under pressure. Charge for the junction rule; energy for the loop rule.

Combining resistors

In series: the same current passes through each, and the p.d.s add.

R = R₁ + R₂ + R₃

In parallel: the same p.d. is across each, and the currents add.

1/R = 1/R₁ + 1/R₂ + 1/R₃

Two checks worth running on any answer:

If your answer breaks either rule, it is wrong, and this catches the frequent slip of forgetting to invert at the end of the parallel calculation.

For two resistors in parallel, R = R₁R₂ / (R₁ + R₂) is quicker. For n identical resistors of resistance R in parallel the combination is R/n.

Worked example. What is the resistance of four identical resistors R connected in parallel, compared with the same four in series?

Parallel: R/4. Series: 4R. The ratio is 1:16.

Internal resistance

A real cell has resistance of its own, and the energy it dissipates internally is not available to the circuit.

E = I(R + r) = V + Ir

where E is the e.m.f., r the internal resistance, R the external resistance and V the terminal potential difference.

So V = E − Ir. The terminal p.d. is always less than the e.m.f. when current flows, and the difference Ir is called the lost volts.

Three consequences that questions rely on:

Worked example. A cell of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor. Find the current and the terminal p.d.

I = E / (R + r) = 1.5 / 3.0 = 0.50 A.

V = E − Ir = 1.5 − 0.50 × 0.50 = 1.25 V.

Plotting V against I gives a straight line of gradient −r with intercept E, and reading r off as a positive gradient is a common error.

The maximum power is delivered to the external resistance when R = r, which is worth knowing even though the derivation is beyond AS.

Potential dividers

Two resistors in series across a supply divide the p.d. in the ratio of their resistances:

V_out = V_in × R₂ / (R₁ + R₂)

The output is taken across R₂. Getting the wrong resistor on the top of the fraction gives the complement of the right answer, so identify which resistor the output is measured across before writing anything.

Worked example. A 12 V supply is across a 4.0 kΩ and an 8.0 kΩ resistor in series. What is the p.d. across the 8.0 kΩ?

V = 12 × 8.0 / 12.0 = 8.0 V.

The larger resistance takes the larger share, which is the sanity check to run.

Sensing circuits

Replace one resistor with a thermistor or an LDR and the output responds to the environment.

The reasoning for these questions is always the same three steps, and it is worth doing them explicitly:

  1. Decide what happens to the resistance of the sensor.
  2. Decide what happens to its share of the total resistance.
  3. Decide what happens to V_out.

For example, a thermistor as the upper resistor with the output across the fixed lower one. As the temperature rises, the thermistor's resistance falls, so it takes a smaller share of the supply, so the p.d. across the fixed resistor rises.

Swap them over and the conclusion reverses. So a question showing four circuits and asking in which one V rises with temperature is asking you to run those three steps on each, and the answer turns on which component the output is across.

A potentiometer is a potential divider with a sliding contact, giving a continuously variable output from zero to the full supply p.d.

Ammeters and voltmeters

A real voltmeter of finite resistance connected across one resistor of a potential divider draws current and lowers the reading below the value calculated. Questions ask why a measured value is lower than the predicted one, and this is the answer.

Common mistakes

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