P6 - Electrical Circuits Quiz

Wed, 22 Jul 2026 00:26:30 +0700

Physics Quiz: P6- Electrical Circuits

Time allowed: 60 minutes Total marks: 50

Section A: Multiple Choice Questions (MCQ) [10 marks]

An ammeter is used to measure which electrical quantity? [1]
A) Potential difference (p.d.)
B) Current
C) Resistance
D) Power
What is the correct circuit position for a fuse in relation to the live wire? [1]
A) In parallel with the live wire
B) In series before the appliance, on the live wire
C) In series after the appliance, on the neutral wire
D) Connected directly to earth

P6 - Electrical Circuits Mark Scheme

Wed, 22 Jul 2026 00:26:10 +0700

Marking Scheme: Electrical Circuits Quiz

This is the marking scheme for quiz: P6 - Electrical Circuits Quiz
Topic guide: P6 - Electrical Circuits

SECTION A: Multiple Choice Answers [10 marks]

Question	Correct Answer	Rationale / Key Point
1	B) Current	Ammeter measures electric current in amperes (A). Connected in series.
2	B) In series before the appliance, on the live wire	Fuse must connect to live side; otherwise if on neutral it provides no protection after switching off.
3	C) Decreases significantly	LDR resistance is inverse to light intensity (hence name). Lower in brightness.
4	B) Parallel circuit only	In parallel, each branch operates independently; series breaks when one component fails.
5	C) 1/R_total = 1/R1 + 1/R2	Standard reciprocal formula for parallel resistors. B is product-over-sum rule (equivalent).
6	B) Microwave oven (13 A)	Higher power appliance draws more current; requires higher rated fuse for safety.
7	A) V_out increases	In potential divider with LDR: if R1 increased, fraction of Vin across it increases.
8	D) Both B and C	Switch on live isolates from high voltage; also protects neutral wire from overheating risks.
9	B) Circuit breakers can be reset after tripping…	Key advantage: no permanent destruction like fuses which require replacement.
10	B) E_total = E1 + E2 + …	Cells in series add algebraically (helping each other configuration).

SECTION B: Short Answer Marking Scheme [30 marks]

11: Circuit Components and Symbols [6 marks]

a) Component Functions [4 marks total, 1 mark each]

Component	Expected Answer(s)	Acceptable Variations
Resistor (Fixed)	Limits/resists current flow; opposes current with constant resistance	“Opposes current”, “Restricts electron flow”, “Dissipates energy as heat”
Diode	Allows current in one direction only; blocks reverse current	“One-way valve for electricity”, “Blocks reverse bias”, “Forward conducts only”
Thermistor (NTC)	Resistance decreases as temperature increases	Variable resistor - NTC = more resistance when cold, less when hot
Transformer	Changes AC voltage levels using mutual induction; steps up/down voltage	“Changes p.d.”, “Voltage transformer”, “Works with AC only”

b) Voltmeter Connected in Parallel [2 marks total]

Mark	Required / Acceptable Point
1 mark	Must state voltmeter measures potential difference across component, NOT breaking current path through it
1 mark	Should mention infinite/very high resistance ideal - connecting in series would break circuit or alter current significantly (“Would block current flow” or “Would introduce significant resistance and reduce circuit performance”)

12: Series and Parallel Circuits [8 marks]

a) Three 1.5V Cells, Single 4Ω Resistor [5 marks total]

Mark	Required / Acceptable Point
1 mark (i)	E_total = 1.5 V × 3 cells = 4.5 V
3 marks (ii)	Current calculation: I = V/R = 4.5/4 = … = 1.125 A (show working) [1.5 + 1.0 + 0.5]
2 marks (iii)	Explain cells add their e.m.f., giving more voltage drive to push current. Series combines forces (“Adding voltages pushes same charge through” - 1 mark; “Greater potential difference drives higher current by Ohm’s Law” - 1 mark)

b) Two 6Ω Resistors in Parallel [5 marks total]

Mark	Required / Acceptable Point
2 marks (i)	Correct reciprocal formula setup: 1/R_total = 1/6 + 1/6 [1 mark]; R_total = 3 Ω or correct calculation [1 mark] Alternatively: R × R / (R + R) = 36/12 = 3 [3 marks for direct answer with working shown]
2 marks (ii)	Currents: I_1 = V/R_1 = 12/6 = 2.0 A. I_2 = 12/6 = 2.0 A (show V=IR rearrangement) [1 mark each] Or: “Each resistor has full supply voltage, so apply Ohm’s Law independently”

13: Potential Dividers and Sensors [8 marks]

a) Calculate V_out for R1=3kΩ, R2=6kΩ, Vin=9V [3 marks]

Mark	Required / Acceptable Point
1 mark	State formula: V_1 = V_in × (R_1/R_1+R_2) [or equivalent]
2 marks (total)	Substitution correct [1 mark]. Final answer: 3.0 V and method clear showing fraction calculation [1 mark for 9 × 3/9 or similar, giving result to appropriate sig figs]

b) Voltmeter Behavior with Sensors [4 marks total]

Mark	Scenario / Expected Answer(s)
2 marks (LDR + Light ↑)	V_out decreases as resistance drops; smaller fraction of Vin dropped across LDR. Alternative wording acceptable: “Light reduces LDR ohms, making divider output less” [1 mark each point]
2 marks (Thermistor + Temp ↓)	V_out increases as NTC thermistor resistance increases in cold; larger voltage fraction drops to it. Alternative: “Cold gives high resistive value and raises divider reading across it” [1 mark each point - 1 for direction, 1 for mechanism explanation]
1 mark (Variable potentiometer)	Sliding contact allows continuous/range adjustment of output voltage ratio without changing fixed values; “Smooth adjustment possible across entire input range from zero to full Vin”

c) Applications and Configurations [4 marks total]

LDR circuit: Dark = High R_LDR → V_out high enough to trigger relay. Light = Low R → V_out low, relay off. Must show understanding of state inversion in logic. Explanation of potential divider setup: Fixed resistor at top connected to Vin, LDR forming bottom leg (to ground) gives maximum output when light is removed. Alternative acceptable configuration explanation [4 marks: 1 mark for dark/light resistance behavior, 2 marks for voltage reading direction, 1 mark for relay activation threshold logic]

14: Electrical Safety and Protection [8 marks]

a) Four Mains Hazards [4 marks - 1 each]

Hazard	Acceptable Answer(s)	Marks
Damaged Insulation	Exposed wires touching live/earth or people; leads to shock or short-circuit fire	1
Overheating Cables	Too many/high-rated currents create heat via I²R losses in thin/coiled cables	1
Damp/Wet Conditions	Water (conductor) reduces skin resistance, allowing much higher current through person at shock time	1
Overloading	Multiple appliances on one wall socket multi-draw; house wires heat up internally from excess mains load; start fires	1

b Fuse Protection Operation [3 marks]

Mark	Required / Acceptable Point
1 mark	Current surges through short circuit live-to-earth fault path exceeding fuse rating
1 mark	Heat generated in thin fused wire (I²R loss) melts the fusible metal element
1 mark	Circuit opens/breaks; appliance disconnected from live supply; prevents earth shock/fire hazard

c Double Insulation [1 mark]

Expected Answer: Fully plastic non-conducting outer case provides secondary protection, so no earth connection needed. Even if internal wiring touches casing, outer layer cannot conduct high voltage to user. Alternative wording acceptable regarding “two layers of insulation prevent contact with live parts”. Note: Avoid mentioning earthing since question asks preference for double insulation over it.

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