Purpose: Test understanding of charge, current, voltage, resistance, Ohm's law, series/parallel circuits, magnetism, and electromagnetism.
Formulae: \( V=IR \) | \( I=Q/t \) | \( P=VI=I^2R=V^2/R \) | \( E=Pt \) | series \( R_T=\sum R_i \), \( I_T=I_1=I_2 \), \( V_T=\sum V_i \) | parallel \( 1/R_T=\sum 1/R_i \), \( V_T=V_1=V_2 \), \( I_T=\sum I_i \) | 1 kWh = \(3.6\times10^6\) J
Q1 — Ohm's law
A resistor draws \(0.50\) A when connected to a \(12\) V battery. What is its resistance?
Answer:
Check
Solution
Solution:
\( R = V/I = 12/0.50 = 24 \) Ω
Q2 — Current from charge
A wire carries \(60\) C of charge in \(30\) s. What is the current?
Answer:
Check
Solution
Solution:
\( I = Q/t = 60/30 = 2.0 \) A
Q3 — Series circuit
Two resistors, \(4.0\) Ω and \(6.0\) Ω, are connected in series across a \(20\) V battery. What is the current through the circuit?
Answer:
Check
Solution
Solution:
\( R_T = 4+6 = 10 \) Ω. \( I = V/R = 20/10 = 2.0 \) A
Q4 — Parallel circuit
Two resistors, \(6.0\) Ω and \(12\) Ω, are connected in parallel. Calculate the equivalent resistance.
Answer:
Check
Solution
Solution:
\( 1/R_T = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 \) → \( R_T = 4.0 \) Ω
Q5 — Power
A heater draws \(8.0\) A from a \(120\) V supply. What is its power?
Answer:
Check
Solution
Solution:
\( P = VI = 120(8.0) = 960 \) W
Q6 — Energy in kWh
A \(1500\) W kettle is on for \(0.20\) hours. How much electrical energy (in kWh) does it use?
Answer:
Check
Solution
Solution:
\( E = Pt = 1.5 \text{ kW} \times 0.20 \text{ h} = 0.30 \) kWh
Q7 — Mixed circuit
A \(12\) V battery is connected to a \(6.0\) Ω resistor in series with a parallel combination of two \(4.0\) Ω resistors. What is the current from the battery?
Answer:
Check
Solution
Solution:
Parallel: \( R_p = (4\times 4)/(4+4) = 2.0 \) Ω
Total: \( R_T = 6.0 + 2.0 = 8.0 \) Ω
\( I = V/R = 12/8 = 1.5 \) A
Q8 — Right-hand rule (concept)
A current flows out of the page through a wire. Using the right-hand rule, the magnetic field around the wire forms loops that go:
Check
Solution
Solution:
Right-hand rule for a straight wire: thumb in the direction of conventional current (out of page), fingers curl counterclockwise as viewed by an observer looking at the page.
Q9 — Cost calculation
A 60-W light bulb runs for 8 hours per day for 30 days. At \(\$0.12\)/kWh, what is the cost?
Answer:
Check
Solution
Solution:
\( E = 0.060 \text{ kW} \times 8 \times 30 = 14.4 \) kWh
Cost \( = 14.4 \times 0.12 = \$1.73 \)
Q10 — EM induction (concept)
A bar magnet is pushed into a coil of wire connected to a galvanometer. What is observed?
Check
Solution
Solution:
Faraday's law: an EMF is induced whenever the magnetic flux through the coil changes. The induced EMF is proportional to the rate of change of flux.
SPH3U — Physics, Grade 11 | Unit 5: Electricity and Magnetism | Ontario Curriculum 2008