Electric Circuits — Science Sage

Electric Current

Electric current is the rate of flow of electric charge. In metals, charge is carried by electrons moving through the conductor.

I = Q / t

Quantity	Symbol	Unit
Current	I	Ampere (A)
Charge	Q	Coulomb (C)
Time	t	second (s)

Conventional vs Electron Current: Conventional current flows from + to − (positive terminal of battery); electrons actually flow from − to +. We always use conventional current direction in circuit diagrams.

Potential Difference (Voltage)

Potential difference (voltage) is the work done per unit charge to move charge between two points. It is the "push" that drives current around a circuit.

V = W / Q

Unit: Volt (V). One volt = one joule per coulomb. A voltmeter is connected in parallel across a component to measure its voltage.

Resistance

Resistance is the opposition to the flow of current. It is caused by collisions between electrons and the lattice of metal ions in the conductor.

R = V / I

Unit: Ohm (Ω). An ammeter is connected in series to measure current. A resistor's resistance depends on: material, length (longer → more R), cross-sectional area (thicker → less R), and temperature.

Ohm's Law

For an ohmic conductor at constant temperature, the current through it is directly proportional to the voltage across it:

V = I × R

The V–I graph of an ohmic conductor is a straight line through the origin. The gradient equals resistance.

Non-ohmic conductors (e.g. light bulb filament) do not produce a straight V–I line, because as temperature increases, resistance increases — the graph curves.

Circuit Symbols

Key symbols used in circuit diagrams:

Cell: short and long parallel lines (short = negative)
Battery: multiple cells in series
Resistor: rectangle
Variable resistor (rheostat): rectangle with an arrow
Lamp (bulb): circle with cross inside
Switch: open gap in line
Ammeter: circle with A — in series
Voltmeter: circle with V — in parallel

Series Circuits

In a series circuit, components are connected end-to-end in a single loop:

The same current flows through all components: I_total = I₁ = I₂ = I₃
Voltages add up: V_total = V₁ + V₂ + V₃
Resistances add up: R_total = R₁ + R₂ + R₃

Example: R₁ = 4 Ω, R₂ = 6 Ω in series → R_total = 10 Ω. With EMF = 20 V: I = V/R = 20/10 = 2 A.

Parallel Circuits

In a parallel circuit, components share the same two nodes:

The same voltage appears across each branch: V_total = V₁ = V₂ = V₃
Currents add up: I_total = I₁ + I₂ + I₃
Reciprocal resistance: 1/R_total = 1/R₁ + 1/R₂ + ...

Note: Adding more parallel branches decreases total resistance and increases total current drawn from the battery.

Power and Energy

P = V × I = I²R = V²/R (unit: Watt, W)

E = P × t = V × I × t (unit: Joule, J)

Electricity billing uses kilowatt-hours (kWh): 1 kWh = 3.6 × 10⁶ J. A 2 kW heater running for 3 hours uses 6 kWh of energy.

EMF and Internal Resistance

A real battery has internal resistance (r) that causes a voltage drop when current flows. The EMF (ε) is the total energy supplied per unit charge by the battery:

V_terminal = ε − Ir

When a high current is drawn (e.g. from a flat battery), the internal resistance drop (Ir) is large, and the terminal voltage drops noticeably.

IEB Extension — Kirchhoff's Laws

Kirchhoff's Current Law (KCL): The sum of currents entering a node equals the sum of currents leaving it. (Conservation of charge)

ΣI_in = ΣI_out

Kirchhoff's Voltage Law (KVL): The algebraic sum of all voltages around any closed loop is zero. (Conservation of energy)

ΣV_loop = 0

Wheatstone Bridge: A balanced bridge has no current through the galvanometer. Balance condition:

R₁/R₂ = R₃/R₄

Used to measure unknown resistances with high precision.

Circuit Simulator

Circuit Type

Values

EMF (V) 12 V

R₁ (Ω) 4 Ω

R₂ (Ω) 6 Ω

Total R

— Ω

Total I

— A

V across R₁

— V

Power (W)

— W

Question 1 — Ohm's Law

A resistor has a voltage of 18 V across it and a current of 3 A through it. What is its resistance?

Correct! R = V/I = 18/3 = 6 Ω.

Use R = V/I = 18 V ÷ 3 A = 6 Ω.

Question 2 — Series Circuit

Three resistors of 3 Ω, 5 Ω, and 2 Ω are connected in series to a 20 V battery. What is the current in the circuit?

Correct! R_total = 3+5+2 = 10 Ω. I = V/R = 20/10 = 2 A. Same current flows through all components in series.

In series, R_total = 3+5+2 = 10 Ω. Then I = V/R = 20/10 = 2 A.

Question 3 — Parallel Circuit

Two resistors, 6 Ω and 12 Ω, are connected in parallel. What is the total (equivalent) resistance?

Correct! 1/R = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 → R = 4 Ω. Parallel resistance is always less than the smallest individual resistor.

1/R_total = 1/6 + 1/12 = 3/12 → R_total = 4 Ω. Never simply add in parallel.

Question 4 — Power

A lamp draws a current of 0.5 A when connected to a 220 V supply. What is the power rating of the lamp?

Correct! P = VI = 220 × 0.5 = 110 W.

P = V × I = 220 V × 0.5 A = 110 W.

Question 5 — EMF & Internal Resistance

A battery has an EMF of 9 V and an internal resistance of 0.5 Ω. When a current of 4 A is drawn, what is the terminal voltage?

Correct! V_terminal = ε − Ir = 9 − (4 × 0.5) = 9 − 2 = 7 V.

V_terminal = ε − Ir = 9 − (4 × 0.5) = 9 − 2 = 7 V. The internal resistance causes a "lost voltage" of Ir = 2 V.

Question 6 — Energy

A 1500 W electric heater runs for 2 hours. How much energy does it consume in joules?

Correct! E = Pt = 1500 W × (2 × 3600 s) = 1500 × 7200 = 10 800 000 J = 10.8 MJ. This equals 3 kWh.

E = P × t. Convert 2 hours to seconds: 2 × 3600 = 7200 s. E = 1500 × 7200 = 10 800 000 J.

Question 7 — Mixed Circuit & Power (Analysis)

A 12 V battery is connected to a 4 Ω resistor (R₁) in series with a parallel combination of R₂ = 6 Ω and R₃ = 12 Ω. Calculate the power dissipated in R₁.

Correct! First find the parallel equivalent of R₂ and R₃: 1/Rₚ = 1/6 + 1/12 = 3/12 → Rₚ = 4 Ω. Total resistance = R₁ + Rₚ = 4 + 4 = 8 Ω. Total current: I = V/R = 12/8 = 1.5 A. Since R₁ is in series, this full 1.5 A flows through it: P = I²R₁ = 1.5² × 4 = 9 W.

Work through it step by step: (1) parallel equivalent of R₂ and R₃: 1/Rₚ = 1/6 + 1/12 = 3/12 → Rₚ = 4 Ω. (2) Total circuit resistance = R₁ + Rₚ = 4 + 4 = 8 Ω. (3) Total (and series) current I = V/R = 12/8 = 1.5 A. (4) Power in R₁: P = I²R₁ = 1.5² × 4 = 9 W.

Question 8 — Internal Resistance & Power (Analysis)

A battery has an EMF of 24 V and an internal resistance of 2 Ω. When connected to an external resistor, a current of 3 A flows. Calculate the power dissipated in the EXTERNAL resistor only (not the total power supplied by the battery).

Correct! Terminal voltage (the voltage actually available to the external resistor) = ε − Ir = 24 − (3×2) = 18 V. Power dissipated in the external resistor = V_terminal × I = 18 × 3 = 54 W. Note this is less than the total power supplied by the EMF (ε×I = 72 W) — the missing 18 W (= I²r = 3²×2) is "lost" heating the battery's own internal resistance.

Don't just use P = εI (that gives the TOTAL power supplied, 72 W). First find the terminal voltage: V_terminal = ε − Ir = 24 − (3×2) = 18 V. This is the voltage actually across the external resistor. Power in the external resistor = V_terminal × I = 18 × 3 = 54 W. The remaining 18 W (I²r) is dissipated inside the battery itself.

IEB Extension Practice IEB ONLY

IEB Question 1 — Kirchhoff's Voltage Law

In a closed loop, a battery of EMF 12 V has internal resistance 1 Ω. It drives current through two external resistors R₁ = 3 Ω and R₂ = 5 Ω in series. Applying KVL, what is the current in the circuit?

Correct! KVL: ε − Ir − IR₁ − IR₂ = 0 → 12 − I(1+3+5) = 0 → I = 12/9 ≈ 1.33 A.

KVL: sum of all voltages around the loop = 0. 12 = I×(1+3+5) = 9I → I = 12/9 ≈ 1.33 A.

IEB Question 2 — Kirchhoff's Current Law

At a junction in a parallel circuit, two branches carry currents of 2.4 A and 1.6 A respectively. A third branch carries an unknown current I₃ away from the junction. A total current of 5.5 A enters the junction. Find I₃.

Correct! KCL: I_in = I₁ + I₂ + I₃ → 5.5 = 2.4 + 1.6 + I₃ → I₃ = 5.5 − 4.0 = 1.5 A.

KCL: total current in = sum of branch currents out. 5.5 = 2.4 + 1.6 + I₃ → I₃ = 1.5 A.

Show all working for calculation questions. Use correct units throughout. Draw clearly labelled circuit diagrams where required.

Question 1 — Circuit Calculations

A circuit contains a 15 V battery connected to three resistors: R₁ = 2 Ω in series with a parallel combination of R₂ = 6 Ω and R₃ = 3 Ω.
(a) Calculate the equivalent resistance of R₂ and R₃ in parallel.
(b) Calculate the total resistance of the circuit.
(c) Calculate the total current from the battery.
(d) Calculate the voltage across the parallel combination.
(e) Calculate the current through R₂.

Question 2 — EMF and Internal Resistance

A battery has an EMF of 6 V. When a 5 Ω resistor is connected, the terminal voltage is 5 V.
(a) Calculate the current in the circuit.
(b) Calculate the internal resistance of the battery.
(c) Calculate the "lost volts" (voltage dropped across internal resistance).
(d) Explain why the terminal voltage of a battery decreases as more current is drawn from it.

Question 3 — Energy and Cost

A household uses the following appliances for the times shown: a 2 kW kettle for 30 min, a 150 W television for 5 hours, and a 60 W lamp for 8 hours.
(a) Calculate the energy consumed by each appliance in kWh.
(b) Calculate the total energy used in joules.
(c) If electricity costs R2.50 per kWh, what is the total cost for these appliances?

Question 4 — Ohmic vs Non-Ohmic

(a) Explain the difference between an ohmic and a non-ohmic conductor. Give one example of each.
(b) Sketch the V–I graph for an ohmic conductor and a light bulb filament on the same axes. Label both curves.
(c) Explain why the light bulb filament is non-ohmic.

Question 5 — Power Relationships

A 60 Ω resistor is connected to a 12 V power supply.
(a) Calculate the current through the resistor.
(b) Calculate the power dissipated using P = V²/R.
(c) Verify your answer using P = I²R.
(d) If the voltage is doubled to 24 V, by what factor does the power increase? Explain using the formula.

Question 6 — V–I Data Table: Ohmic vs Non-Ohmic

A learner sets up a circuit to investigate two conductors: a fixed resistor (Conductor A) and a light bulb filament (Conductor B). For each conductor, the learner varies the voltage and records the current:

Voltage (V)	Current through A (A)	Current through B (A)
2	0.40	0.50
4	0.80	0.72
6	1.20	0.86
8	1.60	0.96

(a) Calculate the resistance of Conductor A at each of the four voltage readings. Show your working.
(b) Calculate the resistance of Conductor B at each of the four voltage readings.
(c) State, with a reason based on your calculations, which conductor is ohmic and which is non-ohmic.
(d) If Conductor A's data were plotted as a V–I graph (V on the y-axis, I on the x-axis), calculate the gradient of the line between (0 A, 0 V) and (1.60 A, 8 V), and state what physical quantity this gradient represents.
(e) Explain, in terms of what happens inside the filament, why Conductor B's resistance changes as the voltage increases.