L3-04. DC Electrical Principles — Kirchhoff's Laws & Circuit Analysis

⚡ DC Electrical Principles — Kirchhoff's Laws & Circuit Analysis

DC circuit analysis is the core skill for designing power supplies, control circuits, sensor interfaces and motor drives. Ohm's law, the power equations and Kirchhoff's two laws let you find every voltage, current and power in any resistive DC network.

V = IRP = IV KCL: ΣI = 0KVL: ΣV = 0 Series / Parallel

2Topics

8Examples

4Calculators

12Practice Qs

⚡ Level 3 context: Kirchhoff's Current Law (KCL) and Voltage Law (KVL) apply to every circuit — simple or complex. They are the formal statement of conservation of charge and conservation of energy in electrical networks.

⚡Ohm's Law & PowerV=IR, P=IV, series/parallel

🔌Kirchhoff's LawsKCL, KVL, circuit analysis

⚙️Applications4 circuit calculators

🎮Game12 randomised questions

📝Practice12 graded questions

⚡

Ohm's Law, Power & Series / Parallel Circuits

Ohm's Law relates voltage, current and resistance. The power equations let you find energy consumed or dissipated. Series and parallel rules extend these to multi-component circuits.

V = IRP = IV = I²R = V²/RSeries Rᵀᵀ = ΣR1/Rᵀᵀ = Σ1/R

🎮Live explorer — Ohm's Law & powerDrag sliders — all values update instantly

Voltage V (V) 12

Resistance R (Ω) 6

Current I

2.000 A

I = V ÷ R

Power P

24.00 W

P = V × I

Energy in 1 hour

86.4 kJ

E = P × t

💡 Double the voltage → double the current → four times the power (P = V²/R).

📋Try it — Ohm's Law & Power CalculatorUpdates as you type

Find

Voltage V (V)

Current I (A)

Resistance R (Ω)

Select what to find and enter values above.

📚 Key vocabulary

Voltage (V): electrical potential difference — the "pressure" that drives current through a circuit. Unit: volts (V).
Current (I): rate of flow of charge. Unit: amperes (A). Conventional current flows from positive to negative terminal.
Resistance (R): opposition to current flow. Unit: ohms (Ω). Depends on material, length and cross-section.
Power (P): rate of energy conversion. P = IV = I²R = V²/R. Unit: watts (W).
EMF (electromotive force): the voltage produced by a source (battery, generator). Drives current around the circuit.
Voltage divider: two resistors in series sharing a voltage: V₁ = Vᵀᵀ × R₁ ÷ (R₁ + R₂).

👁️ Visual aid — series vs parallel: same two resistors (6 Ω and 12 Ω) connected in each configuration

✍️Worked example 1 — series circuit step through it

QuestionA 24 V battery drives current through three series resistors: R₁ = 2 Ω, R₂ = 4 Ω, R₃ = 6 Ω. Find the total resistance, current and voltage across each resistor.

Total RRᵀᵀ = 2 + 4 + 6 = 12 Ω

CurrentI = V ÷ Rᵀᵀ = 24 ÷ 12 = 2 A

V₁V₁ = I × R₁ = 2 × 2 = 4 V

V₂, V₃V₂ = 2×4 = 8 V V₃ = 2×6 = 12 V

Check4 + 8 + 12 = 24 V ✓

Step 1 of 6

✍️Worked example 2 — parallel circuit

QuestionThree resistors R₁ = 6 Ω, R₂ = 12 Ω, R₃ = 4 Ω are connected in parallel across 12 V. Find Rᵀᵀ, total current and branch currents.

1/Rᵀᵀ1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 6/12 = 0.5 → Rᵀᵀ = 2 Ω

Total IIᵀᵀ = 12 ÷ 2 = 6 A

BranchesI₁=12/6=2A I₂=12/12=1A I₃=12/4=3A

Check2 + 1 + 3 = 6 A ✓

✍️Worked example 3 — power dissipation

QuestionA 10 Ω resistor carries 3 A. Find the voltage across it, the power dissipated and the energy consumed in 30 minutes.

VoltageV = IR = 3 × 10 = 30 V

PowerP = I²R = 9 × 10 = 90 W (check: P = IV = 3×30 = 90 W ✓)

EnergyE = P × t = 90 × 1800 = 162 000 J = 162 kJ

AnswerV = 30 V P = 90 W E = 162 kJ

✍️Worked example 4 — voltage divider

QuestionTwo resistors R₁ = 3 kΩ and R₂ = 7 kΩ form a voltage divider across 20 V. Find the voltage across R₂.

FormulaV₂ = Vᵀᵀ × R₂ ÷ (R₁ + R₂)

SubstituteV₂ = 20 × 7000 ÷ (3000 + 7000) = 20 × 0.7

AnswerV₂ = 14 V

🔌

Kirchhoff's Laws & Circuit Analysis

KCL and KVL extend Ohm's Law to complex networks with multiple sources and branches. They are the formal statement of charge and energy conservation in circuits.

KCL: ΣI = 0KVL: ΣV = 0NodeLoopBranch

📜 Key rules

KCL (Kirchhoff's Current Law): sum of currents into a node = sum of currents out. ΣIᴵₙ = ΣIᴼᵁᵀ.

KVL (Kirchhoff's Voltage Law): sum of all voltages around any closed loop = 0. ΣV = 0.

Sign convention: voltage rises (EMF in direction of travel) are positive; voltage drops (IR in direction of current) are negative.

For a loop: +EMF − I×R₁ − I×R₂ = 0 — rearrange to find unknown current.

At a node: if I₁ enters and I₂, I₃ leave, then I₁ = I₂ + I₃.

🎮Live explorer — KVL single loopDrag sliders — voltages and current update

EMF source E (V) 20

Resistor R₁ (Ω) 4

Resistor R₂ (Ω) 6

Current I

2.00 A

E/(R₁+R₂)

V across R₁

8.0 V

I × R₁

V across R₂

12.0 V

I × R₂

KVL check

0.0 V ✓

E−V₁−V₂

💡 KVL: go around the loop adding EMFs (gains) and subtracting voltage drops (I×R). The total must always equal zero.

📋Try it — KVL Single Loop CalculatorUpdates as you type

EMF E (V)

R₁ (Ω)

R₂ (Ω)

Enter EMF and both resistors to solve the loop.

📚 Key vocabulary

Node: a junction point where two or more circuit branches connect. KCL applies at every node.
Branch: a single path between two nodes — contains one or more elements in series.
Loop: any closed path through a circuit. KVL applies to every independent loop.
KCL: Kirchhoff's Current Law — ΣI into a node = ΣI out. Based on conservation of charge.
KVL: Kirchhoff's Voltage Law — ΣV around any closed loop = 0. Based on conservation of energy.
Internal resistance: the resistance inside a real battery or source — causes terminal voltage to drop under load: Vᵀ𝒊𝑟𝑚 = EMF − Ir.

👁️ Visual aid — KCL at a node: I₁ = 6 A enters; I₂ = 4 A and I₃ = 2 A leave. Sum at node = 0.

✍️Worked example 5 — KVL to find unknown current step through it

QuestionA single loop circuit has EMF = 30 V, R₁ = 5 Ω, R₂ = 7 Ω and R₃ = 3 Ω all in series. Use KVL to find the current and each voltage drop.

KVL loop+30 − 5I − 7I − 3I = 0

Collect I30 = I(5 + 7 + 3) = 15I

CurrentI = 30 ÷ 15 = 2 A

DropsV₁=2×5=10V V₂=2×7=14V V₃=2×3=6V

Check10 + 14 + 6 = 30 V ✓

Step 1 of 6

✍️Worked example 6 — KCL at a node

QuestionAt a node, I₁ = 8 A flows in. Two branches leave: R₂ = 4 Ω and R₃ = 12 Ω, both connected to the same voltage V. Find V and each branch current using KCL.

KCLI₁ = I₂ + I₃ → 8 = V/4 + V/12 = V(3/12 + 1/12) = V/3

VoltageV = 8 × 3 = 24 V

CurrentsI₂ = 24/4 = 6 A I₃ = 24/12 = 2 A

Check6 + 2 = 8 A ✓

✍️Worked example 7 — terminal voltage with internal resistance

QuestionA battery has EMF = 12 V and internal resistance r = 0.5 Ω. It supplies 4 A. Find the terminal voltage.

FormulaVᵀ𝒊𝑟𝑚 = EMF − Ir

SubstituteVᵀ𝒊𝑟𝑚 = 12 − 4 × 0.5 = 12 − 2

AnswerVᵀ𝒊𝑟𝑚 = 10 V

✍️Worked example 8 — two-source KVL loop

QuestionA loop contains E₁ = 20 V, E₂ = 8 V (opposing), R₁ = 3 Ω and R₂ = 5 Ω. Find the loop current.

KVL+20 − 8 − 3I − 5I = 0

Collect12 = 8I → I = 12 ÷ 8 = 1.5

AnswerI = 1.5 A

⚙️

Engineering Applications

Four DC circuit calculators for practical design scenarios.

⚡ Cable voltage drop

Find the voltage drop along a cable and the terminal voltage at the load, given cable resistance and current.

V𝑑𝑟𝑜𝑝 = I × R𝑐𝑎𝑏𝑙𝑒

🧮 Calculator

Supply V (V)

Current I (A)

Cable R (Ω)

Enter supply voltage, current and cable resistance.

🔌 Parallel bank — total current

Find the total current drawn from a supply by up to three parallel resistive loads.

Iᵀᵀ = V/R₁ + V/R₂ + V/R₃

🧮 Calculator

Voltage V (V)

R₁ (Ω)

R₂ (Ω)

R₃ (Ω, opt)

Enter voltage and at least two resistors.

📈 Power budget — heater / motor

Calculate current drawn, resistance and energy cost for a given power rating and supply voltage.

I = P ÷ V R = V² ÷ P

🧮 Calculator

Power P (W)

Voltage V (V)

Enter power rating and supply voltage.

🔋 Battery terminal voltage

Find the terminal voltage of a real battery under load, accounting for internal resistance.

Vᵀ𝒊𝑟𝑚 = EMF − I × r

🧮 Calculator

EMF (V)

Current I (A)

Int. resistance r (Ω)

Enter EMF, current and internal resistance.

🎮

DC Circuits Challenge

12 randomised questions — Ohm's law, power and Kirchhoff's laws.

⚡ Quick-fire round

Question 1 / 12 Score: 0 Streak: 0 🔥

Press Start to begin!

📝

Practice Questions

12 graded questions — type your answer to check, then reveal full working.

⚡ Show Ohm's law or Kirchhoff's law used, substitution and every step. State units in every answer.

Q1 · Ohm's law1 mark

A 15 Ω resistor has 3 A flowing through it. Calculate the voltage across it.

V = IR = 3 × 15 = 45 V

Q2 · Ohm's law1 mark

A 60 V supply drives current through a 20 Ω resistor. Find the current.

I = V ÷ R = 60 ÷ 20 = 3 A

Q3 · Power2 marks

A resistor draws 4 A at 50 V. Calculate the power dissipated.

P = IV = 4 × 50 = 200 W

Q4 · Power2 marks

A 25 Ω resistor carries 2 A. Find the power dissipated and the energy used in 10 minutes.

P = I²R = 4×25 = 100 W
E = Pt = 100×600 = 60 000 J = 60 kJ

Q5 · Series circuit2 marks

Three resistors 4 Ω, 6 Ω and 10 Ω are connected in series across 40 V. Find the current and the voltage across the 6 Ω resistor.

Rᵀᵀ = 4+6+10 = 20 Ω
I = 40÷20 = 2 A
V₆ = 2×6 = 12 V

Q6 · Parallel circuit2 marks

Two resistors 8 Ω and 24 Ω are connected in parallel across 24 V. Find Rᵀᵀ and total current.

Rᵀᵀ = 8×24÷32 = 6 Ω
Iᵀᵀ = 24÷6 = 4 A

Q7 · Voltage divider2 marks

A 48 V supply connects to a voltage divider with R₁ = 6 kΩ and R₂ = 10 kΩ. Find the voltage across R₂.

V₂ = 48×10000÷(6000+10000) = 48×0.625 = 30 V

Q8 · KVL2 marks

A single loop: EMF = 36 V, R₁ = 4 Ω, R₂ = 8 Ω. Use KVL to find current and voltage drops.

KVL: 36 − 4I − 8I = 0 → I = 36÷12 = 3 A
V₁ = 12 V V₂ = 3×8 = 24 V Check: 12+24=36 ✓

Q9 · KCL2 marks

At a node: I₁ = 10 A enters; I₂ = 6 A leaves via one branch. Find I₃ in the second branch using KCL.

KCL: I₁ = I₂ + I₃ → 10 = 6 + I₃ → I₃ = 4 A

Q10 · Internal resistance2 marks

A battery (EMF = 9 V, r = 0.6 Ω) supplies 5 A. Find the terminal voltage.

Vᵀ𝒊𝑟𝑚 = EMF − Ir = 9 − 5×0.6 = 9 − 3 = 6 V

Q11 · Combined3 marks

A 30 V supply drives a series circuit: R₁ = 5 Ω, R₂ = 7 Ω. Find (a) current, (b) V across each resistor, (c) power in R₂.

I = 30÷12 = 2.5 A
V₁ = 12.5 V V₂ = 2.5×7 = 17.5 V
P₂ = I²R₂ = 6.25×7 = 43.75 W... or P₂=IV₂=2.5×17.5=43.75 W

Q12 · Two sources KVL3 marks

A loop has E₁ = 24 V, E₂ = 6 V (opposing), R₁ = 6 Ω, R₂ = 3 Ω. Use KVL to find the current.

KVL: 24 − 6 − 6I − 3I = 0
18 = 9I → I = 2 A