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.
SECTION C: Extended/Essay Questions [10 marks]#
15: Thermistor Potential Divider Circuit [4 marks] (Note: Slight discrepancy in total, adjust if 5 instead of 3)#
Assuming 4 mark distribution for this part:
| Point |
Marks |
Acceptable Variation |
| Circuit diagram - LDR or Thermistor at bottom with Vin top to source, output between them connected to ground/earth terminal |
1 |
Proper symbol used (LDR for Light Dependent Resistor if specified) OR thermistor symbol labelled NT C properly |
| Voltage calculations at each temp: Show V_out formula correctly substituted using correct resistance value [3 marks: 2 for math, 1 per calculation to appropriate sigs] |
3 marks combined |
Example working shown acceptable even with minor rounding differences |
Key Expected Values:
- At 6 kΩ thermistor: V_out = 9 × (6/6+3) = 6.0V
- At 1 kΩ thermistor: V_out = 9 × (1/1+3) = 2.25V
16: Applications and Circuit Comparisons [5 marks]#
a) Parallel Domestic Lighting [3 marks]#
| Component |
Expected Answer(s) |
Acceptable Wording |
Marks |
| Resistance 4 lamps parallel: Formula setup showing 1/R_total = 1/120 + etc., or product-over-sum for first two then extend appropriately. Result: 30 Ω (or equivalent calculation) [1 mark] (assuming simple equal resistor case where all are same resistance R, formula gives R/4) |
“R_combined = R/n where n=4” gets the point if justified by showing parallel addition formula properly |
1 |
|
| Current: I = V/R_total = 230/30 = … ≈ 7.6 A (or correct value to 2 sig figs). Show Ohm’s Law rearrangement explicitly [0.5-1 mark for method, remainder for answer accuracy if shown] |
If using 5 lamps scenario instead: I = 230/(4×(1/120)⁻¹) = 230/30 ≈ 7.67 A. Acceptable rounding to 7.7 A [2 marks total] |
2 marks combined for method plus result |
|
| Reason lamps stay lit after one fails: Use circuit independence concept. Each branch connected to live neutral separately. Branch currents continue unchanged as resistance of remaining branches doesn’t change or total supply current changes but not voltage across each lamp |
“Voltage remains constant at 230V because parallel = independent paths”, “One broken filament only opens that branch, loop in other branches still closed” [3 marks split: 1 for concept of independence/splitting, 1 for unaffected others, 1 for constant full mains voltage explanation] |
3 marks combined for conceptual answer with justification |
|
b) Series Lamp Arrangement Problems [5 marks total]#
| Component |
Expected Answer(s) |
Acceptable Wording |
Marks |
| Resistance Calculation: R_total = 4 × 120 Ω = … show working or clearly state result. Must use multiplication for this case as components are equal in series [1 mark] |
Answer: 480 Ω with method shown or formula stated correctly R_series = sum of all Rs. Acceptable: “Adds resistances: 120 + 120 + 120 + 120 = 480Ω” |
1 |
|
| Current Calculation: I = V/R_total = 230/480 = … ≈ 0.479 A (or 0.48 A to 2 sig figs) [1 mark for method/rearrangement, remainder for answer] |
Working: 230 ÷ 480 = 0.4792 A accepted to reasonable precision; “less than 1 Amp” with calculation shown gets partial credit [1.5 marks for full solution or 1 if only showing V÷R formula] |
1.5 marks combined |
|
| Why NOT suitable: Compare parallel vs series performance and safety |
Points required: (i) Current drops significantly in series reducing brightness [0.5]; (ii) If ANY one lamp fails entire circuit goes dark [2]; (iii) Lower voltage per lamp results if resistances differ or supply not maintained; also harder to control independently. Or any of these combined with correct reasoning about why series arrangement makes lighting impractical compared to parallel wiring in homes |
3 marks for comprehensive explanation covering: failure mode (total darkness when one fails); brightness/current reduction due to higher resistance overall; lack of independent switching [Alternative explanations accepted if demonstrating understanding of safety, reliability, or usability] |
|
MARKING SCHEME SUMMARY TABLES#
| Grade |
Marks Range |
% of Total |
Description |
|
| A+ |
46-50 |
92-100% |
Complete answers showing thorough understanding, all calculations correct with proper method shown |
|
| A |
38-45 |
76-90% |
Most answers correct; minor calculation errors or missing points on optional parts |
|
| B |
31-37 |
62-75% |
Major topics covered but with incomplete explanations; formulas stated correctly, some working shown |
|
| C |
24-30 |
48-61% |
Core knowledge demonstrated; basic formulas and definitions present, though detailed reasoning needed improvement |
|
| **D- |
F** |
<24 |
Under 48% |
Attempt attempted but significant gaps in fundamental understanding or calculation errors throughout |
Key Skills Assessed#
- Fundamental Understanding: Knows series/parallel rule differences correctly, current splits/junction conservation
- Calculation Accuracy: Correct application of electrical formulae (Ohm’s Law, parallel resistors, potential dividers) with proper working
- Safety Awareness: Recognizes mains hazards, understands fuse operation and earthing/double-insulation protection mechanisms
- Sensor Knowledge: Comprehends how LDRs and thermistors respond to environmental stimuli; explains circuit configuration for automatic switches
End of Marking Scheme
All marks must be allocated based on demonstrated knowledge rather than memorization. Understanding + calculation method required for full marks on numerical questions.