Internal Resistance of a Source: MDCAT Physics notes
Internal Resistance of a Source for MDCAT: emf versus terminal voltage, V = E − Ir, open circuit, charging a battery, and power delivered to a load.
Emf and internal resistance
The emf $\varepsilon$ of a source is the energy it converts to electrical form per unit charge passing through it. Every real source also has an internal resistance $r$: the resistance of its electrolyte, electrodes or windings.
When a current $I$ flows through an external resistance $R$:
$$\varepsilon = IR + Ir \qquad I = \frac{\varepsilon}{R + r}$$
Terminal potential difference
The voltage across the terminals is the emf minus the drop inside the source:
$$V_t = \varepsilon - Ir$$
- $Ir$ is the "lost volts". The energy $I^2r$ is wasted as heat inside the battery; this is why internal resistance lowers the terminal voltage.
- For a given current, larger $r$ means smaller $V_t$.
- A source with high internal resistance is unsuitable for heavy loads: a large current causes a large internal drop, leaving little voltage at the terminals.
Three situations
| Situation | Relation |
|---|---|
| No current drawn (open circuit) | $V_t = \varepsilon$ |
| Battery supplying current (discharging) | $V_t = \varepsilon - Ir$, so $V_t < \varepsilon$ |
| Battery being charged (current forced in) | $V_t = \varepsilon + Ir$, so $V_t > \varepsilon$ |
A high-resistance voltmeter connected alone across a cell draws almost no current, so it reads very nearly the emf.
Finding internal resistance
$$r = \frac{\varepsilon - V_t}{I}$$
Example: a cell of emf 6 V gives 5.2 V across its terminals when delivering 0.4 A. Then $r = (6 - 5.2)/0.4 = 2\ \Omega$.
Power delivered to the load
The power in the external resistance is
$$P = I^2R = \left(\frac{\varepsilon}{R + r}\right)^2R$$
Worked example: a supply of emf 12 V and internal resistance 1 $\Omega$ feeds a 5 $\Omega$ heater. $I = 12/6 = 2$ A. Power in the heater $= 2^2\times5 = 20$ W; power wasted inside $= 2^2\times1 = 4$ W; total power from the emf $= \varepsilon I = 24$ W. Terminal voltage $= 12 - 2 = 10$ V.
Always use the load resistance for "power delivered to the load", not $R + r$.
Key formulas
- $I = \varepsilon/(R + r)$
- $V_t = \varepsilon - Ir = IR$
- Charging: $V_t = \varepsilon + Ir$
- $r = (\varepsilon - V_t)/I$
- $P_{\text{load}} = I^2R$
Common MDCAT traps
- $V_t$ exceeds the emf only while the battery is being charged.
- $V_t = \varepsilon$ when no current flows (open circuit), not when current is maximum.
- Forgetting to add $r$ to $R$ when finding the current.
- Using $\varepsilon I$ or $I^2(R + r)$ when the question asks only for power in the load.
Quick revision
- Lost volts $= Ir$.
- Internal resistance causes heating inside the source.
- Terminal voltage falls as internal resistance increases.
- Open-circuit terminal voltage equals emf.