Practical circuits and internal resistance
Why the headlights dim
- Start a car and the headlights dim for a moment.
- The starter motor draws a huge current, and the battery's own voltage sags.
- Every real source has some internal resistance.
e.m.f. and p.d.
- e.m.f. $\varepsilon$ — energy each coulomb is given by the source.
- p.d. — energy each coulomb gives up to a component. Both in volts.
The e.m.f. is the energy given to each unit of charge by the ____.
The source (battery, cell, generator) supplies energy to the charge; a component takes it back out (the p.d.).
Internal resistance
- A real source has internal resistance $r$, so some energy is lost inside it.
- The terminal p.d. delivered outside is $V = \varepsilon - Ir$.
A cell of e.m.f. $12\ \text{V}$ and internal resistance $0.50\ \Omega$ drives a current of $2.0\ \text{A}$. What is the terminal p.d.?
$V = \varepsilon - Ir = 12 - 2.0 \times 0.50 = 11\ \text{V}$.
What happens as current changes
- No current ($I = 0$): terminal p.d. equals the e.m.f.
- More current: the terminal p.d. drops (that $Ir$ loss grows).
- Short circuit ($R_{\text{ext}} \to 0$): $I = \dfrac{\varepsilon}{r}$ — a large current.
A cell of e.m.f. $6.0\ \text{V}$ and internal resistance $0.50\ \Omega$ is short-circuited. What current flows?
With $R_{\text{ext}} \to 0$, $I = \dfrac{\varepsilon}{r} = \dfrac{6.0}{0.50} = 12\ \text{A}$.
Measuring the internal resistance
- Vary the external resistor and plot terminal p.d. $V$ against current $I$.
- The line's intercept is $\varepsilon$ and its gradient is $-r$.
On a terminal-p.d.-against-current graph, match each feature.
From $V = \varepsilon - Ir$: at $I = 0$, $V = \varepsilon$ (intercept); the slope is $-r$.
Ammeters and voltmeters
- An ideal ammeter has zero resistance and goes in series.
- An ideal voltmeter has infinite resistance and goes in parallel.
An ideal ammeter has:
An ammeter goes in series and should add no resistance, so an ideal one has zero resistance.
An ideal voltmeter has infinite resistance and is connected in parallel.
Yes — in parallel, with infinite resistance so it draws no current from the component it measures.
You've got it
- terminal p.d. $V = \varepsilon - Ir$ (drops as current rises)
- a $V$–$I$ graph gives $\varepsilon$ (intercept) and $-r$ (gradient)
- ideal ammeter: 0 Ω in series; ideal voltmeter: ∞ Ω in parallel