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UY1: Electromotive Force & Power In Circuits

Model real sources with emf and internal resistance, then compute terminal voltage, delivered power, and internal loss.

  • University Physics Year 1
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Learning objectives

  • Analyse capacitance, resistance, energy transfer, and transient circuit behaviour.
Why this matters + quick links

This page gives the UY1 working model/result for Electromotive Force & Power In Circuits. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

1) At a glance

  • Emf E is energy supplied per unit charge by a source.
  • Real sources have internal resistance r.
  • Key relations (discharging source):
V = E-Ir, P_load = IV, Pᵢₙₜ = I²r
  • Unit reminder: volts and emf are both J C⁻¹.
  • Quick checks: V ≤ E during discharge; V ≥ E during charging (current forced into the source).

Prerequisites: Resistance And Resistivity
Next uses: RC Circuits, R-L Circuit

2) Setup

Use a source of emf E with internal resistance r connected to external load R.

  • Current direction chosen from source positive terminal through load.
  • Loop equation for discharge: E = I(R + r).
  • Terminal voltage across load: V = IR.

Sign convention matters:

  • During discharge: V < E because of internal drop Ir.
  • During charging (current forced into source): terminal voltage can be V = E + Ir.
Common traps (emf vs terminal voltage)
  • Emf E is not “the voltage across the battery in all situations”. The terminal voltage depends on current and internal resistance.
  • Be explicit about the process:
  • Discharge: current leaves the positive terminal, so V = E-Ir.
  • Charge: current enters the positive terminal, so V = E + Ir.
  • Power checks catch sign mistakes: EI must account for both useful load power and internal heating.

3) Core derivation/explanation

From KVL for discharge:

E-IR-Ir = 0 ⇒ I = E/(R + r).

Terminal voltage:

V = IR = E-Ir.

Power balance:

P_source = EI, P_load = IV = I²R, Pᵢₙₜ = I²r.

So

EI = I²R + I²r,

which is energy conservation per unit time.

Efficiency of transfer to the load:

η = P_load/P_source = R/(R + r).

Checks (limiting cases)

  • Open circuit (R → ∞): I → 0 and V → E.
  • Short circuit (R → 0): I → E/r and most power is lost internally as I²r.

4) Worked example(s)

A battery has E = 12.0 V and r = 0.50 Ω, connected to R = 5.0 Ω.

Current:

I = 12.0/(5.0 + 0.50) = 2.18 A.

Terminal voltage:

V = IR = (2.18)(5.0) = 10.9 V.

Power to load:

P_load = I²R = (2.18)²(5.0) = 23.8 W.

Internal heating:

Pᵢₙₜ = I²r = (2.18)²(0.50) = 2.38 W.

Check: EI = (12.0)(2.18) = 26.2 W ≈ P_load + Pᵢₙₜ.

5) Practice set (with hints + answers)

  1. A source has E = 9.0 V, r = 1.0 Ω, load R = 8.0 Ω. Find I.
  2. Using Q1, find terminal voltage V.
  3. For fixed E and r, what happens to current when R increases?

Hints

  • Use I = E/(R + r).
  • Then V = IR or V = E-Ir.
  • Total series resistance rises when R rises.

Answers

  1. I = 1.0 A.
  2. V = 8.0 V.
  3. Current decreases.

6) Summary + next steps

  • Emf is a source property; terminal voltage depends on current and internal resistance.
  • Always separate external useful power from internal loss.
  • KVL plus power balance gives robust sign/units checks.

Next: RC Circuits Previous: Resistance Of A Cylindrical Resistor Back To Electromagnetism