Electricity, Magnetism & Circuits — Problem set, Grade 12
Relate charge, fields, voltage, circuits, induction, stored energy, and electrical power.
Review the key ideas
Electric Charge & Fields
Charge is quantized and conserved; like charges repel and opposites attract with a force given by Coulomb's law F = kq₁q₂/r²; every charge sets up an electric field E = F/q that points away from positive and toward negative charge.
Current, Voltage & Resistance
Current is charge per time; voltage is energy per charge; resistance describes component behavior. For a uniform ohmic wire R=ρL/A under specified conditions.
Circuits & Ohm's Law
Ohm's law V = IR ties voltage, current, and resistance together; series resistances add while parallel resistances give a smaller total (1/R = 1/R₁ + 1/R₂ + …); electrical power is P = IV.
Magnetism & Electromagnetism
Currents produce magnetic fields; forces depend on vector geometry. Changing linked flux induces emf, while a current also requires an appropriate conducting path and energy input.
Electric Potential, Potential Energy & Capacitors
Voltage is energy per charge; capacitors store energy through separated charge.
Electrical Power, Energy & Circuit Design
Compute component power with its own voltage/current and distinguish watts from energy.
Problem 1
Answer and explanation
0.9 N
F = kq₁q₂/r² = (9 × 10⁹)(1 × 10⁻⁶)(1 × 10⁻⁶) ÷ (0.1)² = (9 × 10⁻³) ÷ 0.01 = 0.9 N. Since the charges have the same sign, the force is repulsive.
Problem 2
Answer and explanation
They repel — like charges repel.
Both rods carry the same sign of charge (positive), and the rule for electric charge is that like charges repel while opposite charges attract. So the two positively charged rods push apart.
Problem 3
Answer and explanation
1.5 × 10⁵ N/C (150,000 N/C)
E = F ÷ q = 0.6 ÷ (4 × 10⁻⁶) = 150,000 N/C, or 1.5 × 10⁵ N/C.
Problem 4
Answer and explanation
3 A
I = Q ÷ t = 24 ÷ 8 = 3 A.
Problem 5
Answer and explanation
12 V
V = W ÷ Q = 60 ÷ 5 = 12 V.
Problem 6
Answer and explanation
The longer wire has the greater resistance.
Resistance increases with length, because charge must push through more material and undergo more collisions. Doubling the length roughly doubles the resistance (R is proportional to length), so the longer wire resists current more.
Problem 7
Answer and explanation
3 A
By Ohm's law, I = V ÷ R = 9 ÷ 3 = 3 A.
Problem 8
Answer and explanation
Total resistance = 10 Ω; current = 2 A
In series, resistances add: R = 2 + 3 + 5 = 10 Ω. Then Ohm's law gives the current: I = V ÷ R = 20 ÷ 10 = 2 A (the same current flows through all three).
Problem 9
Answer and explanation
3 Ω
1 ÷ R = 1/6 + 1/6 = 2/6 = 1/3, so R = 3 Ω. For two equal resistors in parallel, the total is always half of one.
Problem 10
Answer and explanation
The meter shows a current whenever the magnet is moving; this is electromagnetic induction.
While the magnet moves, the magnetic field through the coil is changing, which induces a voltage and drives a current (the needle deflects). Pushing the magnet in and pulling it out send current in opposite directions. When the magnet is held still, the field stops changing and the current stops. This production of current by a changing magnetic field is electromagnetic induction, discovered by Faraday.
Problem 11
Answer and explanation
0.72 N
F = BIL = 0.3 × 6 × 0.4 = 0.72 N. This is the force that makes an electric motor's coil turn.
Problem 12
Answer and explanation
24 W
P = IV = 2 × 12 = 24 W.
Problem 13
Answer and explanation
Current = 4 A; power = 960 W
Current: I = V ÷ R = 240 ÷ 60 = 4 A. Power: P = IV = 240 × 4 = 960 W. (You can check with P = V²/R = 240² ÷ 60 = 57,600 ÷ 60 = 960 W.)
Problem 14
Answer and explanation
F=qE=.002N along the field; a negative charge would reverse direction.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 15
Answer and explanation
F=qE gives+0.0010 N east for+2 μC and−0.0015 N, i.e.west, for−3 μC. The field is unchanged in the ideal nonperturbing-test model.
F=qE gives+0.0010 N east for+2 μC and−0.0015 N, i.e.west, for−3 μC.
The field is unchanged in the ideal nonperturbing-test model.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 16
Answer and explanation
I=Q/t=3 A; conventional current direction is positive-charge flow.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 17
Answer and explanation
I=9/12=0.75 A; P=VI=6.75 W; E=Pt=270 J. The stated constant resistance and voltage must persist over the interval.
I=9/12=0.75 A; P=VI=6.75 W; E=Pt=270 J.
The stated constant resistance and voltage must persist over the interval.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 18
Answer and explanation
1/R=1/6+1/6=1/3, soR=3ohm; lower than either branch.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 19
Answer and explanation
Currents are3 A and1 A; total4 A. Req=12/4=3 Ω and P=12×4=48 W. Branch powers36 W and12 W sum consistently.
Currents are3 A and1 A; total4 A.
Req=12/4=3 Ω and P=12×4=48 W.
Branch powers36 W and12 W sum consistently.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 20
Answer and explanation
A changing magnetic flux, through changing field, area, or orientation; a moving magnet can supply that change.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 21
Answer and explanation
F=|q|vB=0.012 N. Parallel motion gives sin0=0 and zero magnetic-force magnitude in this model; electric forces would need a separate account.
F=|q|vB=0.012 N.
Parallel motion gives sin0=0 and zero magnetic-force magnitude in this model; electric forces would need a separate account.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 22
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
0.01 C
Q=CV=.010.
Problem 23
Answer and explanation
Voltage grows during charging; average charging voltage is half final voltage for a linear capacitor.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 24
Answer and explanation
Q=400 μC; U=½×50×10⁻⁶×64=0.0016 J=1.6 mJ. At16 V, U=6.4 mJ, four times as much. Actual ratings must be separately respected.
Q=400 μC; U=½×50×10⁻⁶×64=0.0016 J=1.6 mJ.
At16 V, U=6.4 mJ, four times as much.
Actual ratings must be separately respected.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Problem 25
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
3000 J
E=Pt.
Problem 26
Answer and explanation
Very low external resistance permits large current; internal resistance then dissipates substantial I²r heat.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 27
Answer and explanation
Saving=(0.015−0.009)×2500=15 kWh=5.40×10⁷ J/year. Production, transport, replacement lifespan or disposal is outside this operational-only result.
Saving=(0.015−0.009)×2500=15 kWh=5.40×10⁷ J/year.
Production, transport, replacement lifespan or disposal is outside this operational-only result.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.