Circuit Spark Quiz
Circuit Spark Quiz
Physics: Electricity & Magnetism Assessment
Student Name:
Date:
Ohm's Law
Series/Parallel
25 Points
Part 1: Schematic Symbols
Match the component with its symbol by writing the letter in the box.
1. Resistor
2. Battery (Multi-cell)
3. Ammeter
4. Open Switch
A A
B
C
D
Part 2: Ohm's Law Essentials
Show your work for each calculation. Formula: V = I × R. Include units.
5. A lightbulb has a resistance of 240 Ω. If it is connected to a 120 V source, find the current.
6. A heating element draws 12 A of current. If its resistance is 10 Ω, find the voltage source.
7. A smartphone charger runs at 5 V and draws 2.1 A. Find the resistance (round to 2 decimals).
Part 3: Circuit Topology
8. True or False: In a series circuit, if one bulb burns out, all other bulbs will also go out.
True
False
9. Calculate total resistance (Req) for each scenario:
Series
Resistors R1 = 15 Ω and R2 = 30 Ω in series.
Parallel
Resistors R1 = 20 Ω and R2 = 20 Ω in parallel.
10. Explain one major advantage of wiring a house in parallel rather than series.
Part 4: Technical Analysis
A circuit has a 12 V battery and three identical 6 Ω resistors connected in series.
a) What is the total current in this circuit?
b) What is the voltage drop across each resistor?
R1 R2 R3 12V SOURCE
Schematic Drawing 1.A
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Circuit Spark Answer Key
Circuit Spark Answer Key
Teacher Reference Guide // Final Version 1.6
25 POINTS TOTAL
Part 1: Schematic Symbols (4 pts)
1. Resistor: C
2. Battery: D
3. Ammeter: A
4. Open Switch: B
Part 2: Ohm's Law Essentials (9 pts)
5. Current calculation (I = V / R):
I = 120V / 240Ω = 0.5 A
6. Voltage calculation (V = I x R):
V = 12A x 10Ω = 120 V
7. Resistance calculation (R = V / I):
R = 5V / 2.1A ≈ 2.38 Ω
Part 3: Circuit Topology (7 pts)
8. Correct Choice: TRUE
9a. Series (R1+R2):
45 Ω
9b. Parallel (1/Rt):
10 Ω
10. House Wiring Key Points:
Full credit for: (1) Independent operation (one bulb doesn't break the whole loop). (2) All devices receive full source voltage.
Part 4: Diagram Analysis (5 pts)
a) Total Current:
Req = 18 Ω
0.67 A
I = 12V / 18Ω
b) Voltage Drop:
Uniform Load
4.0 V
V = 0.67A x 6Ω
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