Skip to content
Educora
EN
AzərbaycancaEnglishРусскийTürkçe
My learningEducora StudioSettings
AdvancedGrades 10–1120 min32 / 44

Circuits: EMF, power and Kirchhoff's rules

Learn about the EMF and internal resistance of a source, Ohm's law for a complete circuit, electrical power and Kirchhoff's rules for complex circuits.

Check yourself
In this lesson you will learn
  • Use Ohm's law for a complete circuit to find the current and terminal voltage
  • Find the work, power and heat produced by a current
  • Solve branched circuits with Kirchhoff's rules

When a car engine is started, the headlights dim for a moment. Why, if the battery is still the same battery? Because the source itself has resistance. In this lesson you will learn how to deal with real power sources and how to calculate any complex circuit.

EMF and internal resistance

Definition
Electromotive force (EMF, ℰ)

The work done by non-electrostatic forces inside the source per unit charge moved through it. Its unit is the volt. Despite its name, EMF is not a force but an energy per charge.

I = ℰ / (R + r), U = ℰ − I · r = I · RI = ℰ / (R + r), U = ℰ − I · r = I · R
where:
  • ℰEMF of the source, in V
  • Rexternal (load) resistance, in Ω
  • rinternal resistance of the source, in Ω
  • Uterminal voltage of the source, in V

Ohm's law for a complete circuit. With R = 0 you get the short-circuit current I = ℰ / r.

Example 1: a battery and a lamp

A lamp of 5.5 Ω is connected to a battery with EMF 12 V and internal resistance 0.5 Ω. Find the current, the terminal voltage and the short-circuit current.

Show solution
I = 12 / (5.5 + 0.5) = 2 A.
U = I · R = 2 · 5.5 = 11 V (1 V is “lost” inside the source).
Short circuit: I = 12 / 0.5 = 24 A — the wires get hot and there is a fire risk.
Example 2: measuring EMF and r

With a 4 Ω resistor connected to a source the current is 1 A; with a 9 Ω resistor it is 0.5 A. Find the EMF and the internal resistance.

Show solution
The EMF is the same in both cases, so we write two equations: ℰ = I₁ · (R₁ + r) = I₂ · (R₂ + r).
1 · (4 + r) = 0.5 · (9 + r) → 4 + r = 4.5 + 0.5r → r = 1 Ω.
ℰ = 1 · (4 + 1) = 5 V.
Interactive
Loading simulation…
For a real source, use R + r instead of R: for example, with ℰ = 12 V and R + r = 6 Ω the current is 2 A. See how sharply the current grows as the resistance drops.

Work and power of a current

P = U · I = I² · R = U² / R, Q = I² · R · tP = U · I = I² · R = U² / R, Q = I² · R · t
where:
  • Pelectrical power, in W
  • Qheat produced in the conductor in time t, in J (Joule's law)

Electricity meters measure energy in kilowatt-hours: 1 kW · h = 3.6 · 10⁶ J.

Example 3: an electric kettle

A 2 kW kettle designed for 220 V runs for 5 minutes. Find the current, the resistance of the heating element and the energy used.

Show solution
I = P / U = 2000 / 220 ≈ 9.1 A.
R = U² / P = 220² / 2000 = 24.2 Ω.
Energy used: E = P · t = 2000 · 300 = 600 000 J = 600 kJ ≈ 0.17 kW · h.

The power delivered to the external circuit is P = ℰ² · R / (R + r)². It is greatest when R = r, and then Pmax = ℰ² / (4r). Check it on the graph below.

Interactive
Loading simulation…
x is the load resistance R (Ω), y the power delivered to it (W); u is the EMF and r the internal resistance. The peak is always at x = r: for ℰ = 12 V and r = 2 Ω, Pmax = 18 W.

Kirchhoff's rules

  1. Junction rule: the total current flowing into a junction equals the total current flowing out (charge does not pile up). ∑Iin = ∑Iout.
  2. Loop rule: around any closed loop the algebraic sum of the EMFs equals the algebraic sum of the voltage drops (I · R). ∑ℰ = ∑I · R.
Example 4: a circuit with two sources

Three branches connect points A and B: 1) a 10 V source ℰ₁ with R₁ = 2 Ω; 2) a 4 V source ℰ₂ with R₂ = 2 Ω; 3) R₃ = 4 Ω. Both sources have their “+” terminal towards A; internal resistances are negligible. Find the currents in the branches.

Show solution
Choose currents from the sources towards A (I₁, I₂) and through R₃ from A to B (I₃).
Junction A: I₃ = I₁ + I₂.
Loop 1 (ℰ₁, R₁, R₃): 10 = 2I₁ + 4I₃.
Loop 2 (ℰ₂, R₂, R₃): 4 = 2I₂ + 4I₃.
So I₁ = 5 − 2I₃, I₂ = 2 − 2I₃, I₃ = 7 − 4I₃ → I₃ = 1.4 A, I₁ = 2.2 A, I₂ = −0.8 A.
The minus sign means the current through ℰ₂ flows the other way: the 4 V source is being charged.

Key points

  • Ohm's law for a complete circuit: I = ℰ / (R + r); terminal voltage U = ℰ − I · r.
  • The short-circuit current I = ℰ / r is very large and dangerous.
  • Power P = U · I = I² · R = U² / R; the load gets maximum power when R = r.
  • Kirchhoff's rules: at a junction ∑Iin = ∑Iout; around a loop ∑ℰ = ∑I · R.

Check yourself

10 questions. Every correct answer earns XP.

1 / 10
An 8 Ω resistor is connected to a battery with EMF 9 V and internal resistance 1 Ω. What is the current?