Gr 11 · Electricity · Deep Study

Understanding Circuits

Electricity is about how charge moves and how it gives up energy. This guide takes you from Ohm's Law through series and parallel circuits, power calculations, and the basics of magnetism — with a worked circuit at every step.

5 sections Worked examples with steps Checkpoint questions Intuition-first explanations
1
Ohm's Law
Voltage is the push, resistance is the opposition, current is what flows

Ohm's Law V = IR is the foundation of all circuit analysis. Memorise the triangle: cover what you want to find, multiply or divide what remains.

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Why does V = IR make sense?
Think of a water pipe. Voltage is the water pressure (the push). Resistance is how narrow the pipe is (the opposition). Current is how much water flows per second. Double the pressure with the same pipe — double the flow. Make the pipe narrower — less flow. V = IR captures exactly this: current is proportional to voltage and inversely proportional to resistance.
Ohm's Law — three forms
V = IR     I = V/R     R = V/I
Ohmic device: V-I graph is a straight line through the origin. Slope = R.
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Key facts:
V = voltage (Volts) · I = current (Amperes) · R = resistance (Ohms, Ω)
An ohmic device obeys V = IR at all values — V-I graph is linear through the origin.
A non-ohmic device (e.g. diode, light bulb filament) has a curved V-I graph — R changes.
Ohm's Law Triangle + V-I Graph V I R Cover V → I × R Cover I → V / R Cover R → V / I Cover the unknown — multiply or divide what remains V I Voltage → Current → Ohmic (linear) Non-ohmic (bulb)
Left: The Ohm's Law triangle — cover the unknown quantity to read the formula. Right: V-I graph for an ohmic device (straight line, slope = R) vs a non-ohmic device (curved, R varies).
Easy
Find current from V and R
R = 40 Ω, V = 120 V. Find the current.
Show solution
1
Identify what is known and what to find

Known: V = 120 V, R = 40 Ω. Find: I.

2
Apply I = V/R
I = V / R = 120 / 40 = 3 A
Answer: I = 3 A
Intermediate
Bulb resistance and power at two voltages
A bulb draws 0.5 A from a 120 V supply. Find R, the current at 100 V (assuming R constant), and the power at 120 V.
Show solution
1
Find resistance
R = V / I = 120 / 0.5 = 240 Ω
2
Current at 100 V (R assumed constant)
I = V / R = 100 / 240 = 0.417 A
3
Power at 120 V
P = VI = 120 × 0.5 = 60 W
Answer: R = 240 Ω · I at 100 V ≈ 0.417 A · P at 120 V = 60 W
Checkpoint 1
a) V = 9 V, R = 180 Ω — find I.
b) I = 2 A, R = 15 Ω — find V.
c) A device draws 3 A at 240 V — find R. Is the device ohmic if it draws 4 A at 240 V when hot?

a) I = V/R = 9/180 = 0.05 A

b) V = IR = 2 × 15 = 30 V

c) R = V/I = 240/3 = 80 Ω. If it draws 4 A at the same 240 V, R = 240/4 = 60 Ω — the resistance changed, so the device is non-ohmic.

2
Series Circuits
One path — current is identical everywhere, voltage divides

In a series circuit, current has only one path. Every coulomb of charge passes through every component — so current is identical everywhere. Voltage divides in proportion to resistance.

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Why is current the same everywhere in series?
Think of a single-lane road with traffic lights. Every car that passes the first light must also pass the second and third — no car can disappear or appear in between. Charge in a series circuit is the same: whatever flows through one resistor flows through all of them. There is nowhere else for it to go.
Series rules
I same · Rₜ = R₁+R₂+… · Vₜ = V₁+V₂+… · V₁/V₂ = R₁/R₂
Series Circuit — Same Current, Voltage Splits + Vₜ R₁ V₁ R₂ V₂ R₃ V₃ I (same everywhere)
A series circuit: one loop, three resistors. Current I is identical everywhere. Voltage drops V₁, V₂, V₃ across each resistor sum to the total supply voltage Vₜ.
Easy
Two resistors in series
12 V battery, R₁ = 4 Ω, R₂ = 8 Ω in series. Find Rₜ, I, V₁, V₂.
Show solution
1
Total resistance
Rₜ = R₁ + R₂ = 4 + 8 = 12 Ω
2
Current (same everywhere)
I = V / Rₜ = 12 / 12 = 1 A
3
Individual voltage drops
V₁ = I × R₁ = 1 × 4 = 4 V
V₂ = I × R₂ = 1 × 8 = 8 V
4
Check: voltages must sum to supply
4 + 8 = 12 V ✓
Answer: Rₜ = 12 Ω · I = 1 A · V₁ = 4 V · V₂ = 8 V
Intermediate
Three resistors — find power in largest
6 Ω, 10 Ω, 14 Ω in series across 90 V. Find Rₜ, I, each voltage drop, and power in the 14 Ω resistor.
Show solution
1
Total resistance
Rₜ = 6 + 10 + 14 = 30 Ω
2
Current
I = V / Rₜ = 90 / 30 = 3 A
3
Individual voltage drops
V₆ = 3 × 6 = 18 V · V₁₀ = 3 × 10 = 30 V · V₁₄ = 3 × 14 = 42 V
Check: 18 + 30 + 42 = 90 V ✓
4
Power in 14 Ω (use P = I²R — same current)
P = I² × R = 3² × 14 = 9 × 14 = 126 W
Answer: Rₜ = 30 Ω · I = 3 A · V₆=18 V, V₁₀=30 V, V₁₄=42 V · P₁₄ = 126 W
Checkpoint 2
a) 24 V supply, R₁ = 3 Ω, R₂ = 5 Ω in series — find I, V₁, V₂.
b) In a series circuit, one bulb burns out (open circuit). What happens to the other bulbs?
c) In a series circuit, which resistor gets the largest voltage drop?

a) Rₜ = 3 + 5 = 8 Ω · I = 24/8 = 3 A · V₁ = 3×3 = 9 V · V₂ = 3×5 = 15 V · Check: 9+15=24 ✓

b) All other bulbs go dark. An open circuit breaks the single path, so no current flows anywhere in the loop.

c) The largest resistor gets the largest voltage drop. (V = IR — same I for all, so V is proportional to R.)

3
Parallel Circuits
Multiple paths — voltage is identical, current divides

In a parallel circuit, each branch gets the full supply voltage. Adding more branches gives current more paths — total resistance decreases, total current increases. This is why household outlets are wired in parallel.

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Why does adding parallel branches lower total resistance?
Think of multiple lanes opening on a highway. Each new lane adds another path for traffic — the total flow increases for the same driving pressure. More paths = less overall resistance. In the extreme: if you connect a wire directly across a battery (a short circuit), resistance approaches zero and current becomes enormous.
Parallel rules
V same · 1/Rₜ = 1/R₁+1/R₂+… · Iₜ = I₁+I₂+… · Rₜ < any branch R
Parallel Circuit — Same Voltage, Current Splits + V Iₜ R₁ I₁ R₂ I₂ R₃ I₃ V (same across all branches) Iₜ = I₁ + I₂ + I₃   |   1/Rₜ = 1/R₁ + 1/R₂ + 1/R₃
A parallel circuit: each branch gets the full supply voltage V. Branch currents I₁, I₂, I₃ are independent — one branch opening does not affect the others. Total current Iₜ = I₁ + I₂ + I₃.
Easy
Two resistors in parallel
24 V, R₁ = 8 Ω, R₂ = 12 Ω in parallel. Find Rₜ, I₁, I₂, Iₜ.
Show solution
1
Total resistance (reciprocal formula)
1/Rₜ = 1/8 + 1/12 = 3/24 + 2/24 = 5/24
Rₜ = 24/5 = 4.8 Ω
2
Branch currents (V same = 24 V across each branch)
I₁ = 24/8 = 3 A · I₂ = 24/12 = 2 A
3
Total current
Iₜ = I₁ + I₂ = 3 + 2 = 5 A
Check: Iₜ = V/Rₜ = 24/4.8 = 5 A ✓
Answer: Rₜ = 4.8 Ω · I₁ = 3 A · I₂ = 2 A · Iₜ = 5 A
Intermediate
Three branches in parallel
R₁ = 6 Ω, R₂ = 12 Ω, R₃ = 4 Ω in parallel across 48 V. Find Rₜ, Iₜ, each branch current.
Show solution
1
Total resistance
1/Rₜ = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 6/12 = 1/2
Rₜ = 2 Ω
2
Total current
Iₜ = V / Rₜ = 48 / 2 = 24 A
3
Branch currents (each branch sees full 48 V)
I₁ = 48/6 = 8 A · I₂ = 48/12 = 4 A · I₃ = 48/4 = 12 A
Check: 8 + 4 + 12 = 24 A ✓
Answer: Rₜ = 2 Ω · Iₜ = 24 A · I₁=8 A, I₂=4 A, I₃=12 A
Checkpoint 3
a) 18 V, R₁ = 9 Ω, R₂ = 6 Ω in parallel — find Rₜ, I₁, I₂, Iₜ.
b) In a parallel circuit, one branch is disconnected. What happens to the voltage and current in the other branches?
c) Why is home wiring parallel rather than series?

a) 1/Rₜ = 1/9 + 1/6 = 2/18 + 3/18 = 5/18 → Rₜ = 18/5 = 3.6 Ω
I₁ = 18/9 = 2 A · I₂ = 18/6 = 3 A · Iₜ = 2+3 = 5 A

b) The other branches are unaffected. Voltage across them remains 18 V; current in each is unchanged. Only total current decreases.

c) Each appliance needs the full supply voltage to operate correctly, and must be switchable independently. Parallel wiring provides both: every outlet gets 120 V and switching one off does not affect the others.

4
Power & Energy
How fast energy is converted — and what it costs

Power tells you how fast a device uses energy. The three power formulas are all equivalent — choose the one that matches what you know. Energy is just power times time; understanding kWh lets you calculate electricity bills.

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Why are there three power formulas?
P = VI is the definition. Substituting V = IR gives P = (IR)I = I²R. Substituting I = V/R gives P = V(V/R) = V²/R. All three are equivalent — they just use different combinations of knowns. The key: use P = I²R when components share the same current (series), and P = V²/R when they share the same voltage (parallel).
Power and energy
P = VI = I²R = V²/R     E = Pt     1 kWh = 3.6×10⁶ J
Power Formulas — Choose Based on What You Know KNOW V and I P = VI Use when V and I are given KNOW I and R P = I²R Use in series (same I) KNOW V and R P = V²/R Use in parallel (same V) = = E = P × t — Energy grows linearly with time E = Pt t = 0 t (seconds) t = max 1 kWh = 3.6 × 10⁶ J  |  E(kWh) = P(kW) × t(h) Tip: double the power → double the energy for the same time; double the time → double the energy.
Three equivalent power formulas — pick the one that matches what you know. The energy bar shows that energy accumulates linearly: E = Pt.
Easy
Energy used by a light bulb
A 60 W bulb runs for 4 hours. Find energy in Joules and in kWh.
Show solution
1
Energy in Joules (convert hours to seconds)
E = Pt = 60 × (4 × 3600) = 60 × 14400 = 864,000 J = 864 kJ
2
Energy in kWh (simpler for billing)
P = 60 W = 0.06 kW
E = 0.06 kW × 4 h = 0.24 kWh
Answer: E = 864,000 J = 0.24 kWh
Intermediate
Power in series vs parallel — same resistors
R₁ = 10 Ω, R₂ = 15 Ω. First in series across 50 V, then in parallel across 50 V. Compare total power.
Show solution
1
Series: find I, then use P = I²R
Rₜ = 25 Ω · I = 50/25 = 2 A
P₁ = I²R₁ = 4×10 = 40 W · P₂ = I²R₂ = 4×15 = 60 W
Pₜ = 100 W · Check: P = VI = 50×2 = 100 W ✓
2
Parallel: each branch has full 50 V — use P = V²/R
P₁ = V²/R₁ = 2500/10 = 250 W
P₂ = V²/R₂ = 2500/15 ≈ 166.7 W
Pₜ ≈ 416.7 W
Answer: Series Pₜ = 100 W · Parallel Pₜ ≈ 416.7 W — parallel dissipates far more power
Checkpoint 4
a) A 1500 W heater runs 2 hours per day. At $0.12/kWh, what is the daily electricity cost?
b) A 100 Ω resistor carries 5 A. Find P using P = I²R.
c) The same 100 Ω resistor is connected across a 500 V supply. Find P using P = V²/R. Do the two answers match?

a) P = 1500 W = 1.5 kW · E = 1.5 × 2 = 3 kWh/day · Cost = 3 × $0.12 = $0.36/day

b) P = I²R = 5² × 100 = 25 × 100 = 2500 W

c) P = V²/R = 500²/100 = 250,000/100 = 2500 W — yes, both formulas give the same answer, as they must (same resistor, V = IR = 5×100 = 500 V).

5
Magnetism
Moving charges create fields — fields exert forces on moving charges

Moving charges create magnetic fields, and magnetic fields exert forces on moving charges. The force is always perpendicular to both the velocity and the field — which is why it causes circular motion, not acceleration along the direction of travel.

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Why is the magnetic force always perpendicular to motion?
A force perpendicular to velocity changes direction but not speed. This is why a charged particle in a uniform magnetic field moves in a circle: the force constantly steers it sideways but never speeds it up or slows it down. It is similar to a ball on a string — the tension is always perpendicular to velocity, so the ball moves in a circle at constant speed.
Magnetic force formulas
F = qvB sinθ   (on a charge)     F = BIL sinθ   (on a wire)
θ = angle between velocity (or current) and B field direction
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Right-hand rules:
Field around a wire: thumb points in direction of conventional current; fingers curl in the direction of the B field.
Force on a wire (motor rule): point fingers in direction of current, curl toward B; thumb gives the direction of the force.
The force is always perpendicular to both the current (or velocity) and B.
Magnetic Field Around a Current-Carrying Wire Current INTO page B Current OUT OF page B Right-hand rule: Thumb points in the direction of conventional current → fingers curl in the direction of B.
Left: current going into the page — field circles clockwise. Right: current coming out of the page — field circles anti-clockwise. Field strength decreases with distance from the wire.
Force on a Current in a Magnetic Field: F = BIL sinθ × × × × × × × × × × × × × × × B (into page) I → F (upward) F = BIL sinθ  (max when θ = 90°: wire ⊥ field) I right × B into page → F upward (right-hand rule / motor rule)
A horizontal wire carrying current I to the right sits in a field B directed into the page. By the right-hand rule (motor rule), the force F = BIL sinθ acts upward — perpendicular to both I and B.
Easy
Force on a moving proton
A proton (q = 1.6 × 10⁻¹⁹ C) moves at 3 × 10⁶ m/s perpendicular to B = 0.5 T. Find the magnetic force.
Show solution
1
Identify: perpendicular means θ = 90°, so sinθ = 1
F = qvB sin90° = qvB
2
Substitute values
F = 1.6×10⁻¹⁹ × 3×10⁶ × 0.5
F = 1.6 × 3 × 0.5 × 10⁻¹⁹⁺⁶
F = 2.4 × 10⁻¹³ N
Answer: F = 2.4 × 10⁻¹³ N (perpendicular to both v and B)
Intermediate
Force on a current-carrying wire at an angle
Wire L = 0.3 m, I = 8 A, B = 0.4 T, θ = 60° to wire. Find F. Then find F if θ = 90°.
Show solution
1
Force at θ = 60°
F = BIL sinθ = 0.4 × 8 × 0.3 × sin60°
sin60° = 0.866
F = 0.4 × 8 × 0.3 × 0.866 = 0.832 N
2
Force at θ = 90° (wire perpendicular to B — maximum)
F = BIL sin90° = BIL = 0.4 × 8 × 0.3 = 0.96 N
Answer: F at 60° = 0.832 N · F at 90° = 0.96 N (maximum possible)
Checkpoint 5
a) An electron (q = 1.6 × 10⁻¹⁹ C) moves at 2 × 10⁵ m/s perpendicular to B = 0.2 T. Find F.
b) A wire 0.5 m long carries I = 10 A in a field B = 0.3 T, at θ = 30°. Find F.
c) If a charge moves parallel to a magnetic field (θ = 0°), what is the magnetic force on it?

a) F = qvB sin90° = 1.6×10⁻¹⁹ × 2×10⁵ × 0.2 = 6.4 × 10⁻¹⁵ N

b) F = BIL sinθ = 0.3 × 10 × 0.5 × sin30° = 0.3 × 10 × 0.5 × 0.5 = 0.75 N

c) F = qvB sin0° = qvB × 0 = 0 N. There is no magnetic force on a charge moving parallel to the field. The force only exists when there is a component of velocity perpendicular to B.