Series and parallel circuits, V = IR, I–V characteristics, LDRs and thermistors – IGCSE Physics exam questions with mark scheme

1Grade 4–55 marks

The diagram shows a series circuit.

(a) Calculate the total resistance of the circuit. [1]

(b) Calculate the reading on the ammeter. [2]

(a) 12 Ω(1)
(b) $I = \dfrac{V}{R} = \dfrac{6.0}{12}$ allow ecf from (a)(1)
(b) 0.50 A(1)
(c) $V = IR = 0.50 \times 8.0$ allow ecf from (b)(1)
(c) 4.0 V(1)
Answer: (a) 12 Ω (b) 0.50 A (c) 4.0 V
Examiner tip: In series the current is the same everywhere, so use the circuit current with the 8.0 Ω resistance only.

2Grade 6–75 marks

The diagram shows two resistors connected in parallel to a 12 V supply.

(a) Calculate the reading on the ammeter. [3]

(b) Use your answer to (a) to calculate the combined resistance of the two resistors. State how this compares with the resistance of each resistor. [2]

(a) $I = V/R$ used for one branch, e.g. $12 / 6.0 = 2.0$ A(1)
(a) other branch $12 / 3.0 = 4.0$ A(1)
(a) 6.0 A (sum of branch currents) allow full marks for a correct alternative method, e.g. combined resistance 2.0 Ω then 12 / 2.0(1)
(b) $R = V/I = 12 / 6.0 = 2.0$ Ω allow ecf from (a)(1)
(b) (combined resistance is) less than either resistor / less than 3.0 Ω allow ecf consistent with their value(1)
Answer: (a) 6.0 A (b) 2.0 Ω, which is less than either resistor
Examiner tip: Each branch of a parallel circuit has the full 12 V across it — do not share the voltage between the resistors.

3Grade 6–75 marks

The graph shows how the current in a filament lamp varies with the voltage across it.

(a) Use the graph to calculate the resistance of the lamp at 2.0 V and at 6.0 V. [3]

(b) Explain why the resistance of the lamp changes as the voltage increases. [2]

(a) $R = V/I$ used with values read from the graph (0.20 A at 2.0 V; 0.30 A at 6.0 V)(1)
(a) 10 Ω (at 2.0 V)(1)
(a) 20 Ω (at 6.0 V)(1)
(b) (as the current increases) the filament gets hotter / its temperature increases(1)
(b) ions / atoms (in the metal) vibrate more, so electrons collide with them more often (so resistance increases) allow so it is harder for electrons to flow through the filament; allow 'the resistance of a metal increases as its temperature increases'; ignore 'more collisions' with no reference to ions, atoms or electrons(1)
Answer: (a) 10 Ω and 20 Ω (b) filament heats up; ions vibrate more, so more collisions with electrons
Examiner tip: Resistance is V ÷ I at a point, not the gradient of the curve — read both values carefully from the grid.

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