Topic 11.712 min read
Terminal Potential Difference
Terminal potential difference is measured across a source's terminals.
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The emf of a cell is the energy it supplies per coulomb. Terminal potential difference is the energy per coulomb available across its external terminals.
A real cell resists charge flow inside itself. When current flows, part of the supplied energy is converted to heat inside the cell.
The internal drop Ir grows when current grows, so a loaded cell can show a smaller voltage than its open-circuit value.
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Terminal potential difference is the potential difference measured across the terminals of a source. For a real source it depends on both its emf and the voltage change within its internal resistance.
Ideal and real sources
An ideal source has zero internal resistance and maintains V=ε for every allowed current. A real cell contains electrolyte and electrodes that oppose charge motion, represented by an internal resistance r in series with an ideal emf ε.
| Quantity | Meaning |
|---|---|
| Emf ε | Energy supplied by the source per unit charge |
| Terminal p.d. V | Energy transferred per unit charge in the external circuit |
| Internal drop Ir | Energy per unit charge converted inside the source |
Energy accounting per unit charge
As one coulomb completes the circuit, the source supplies ε joules. While the cell discharges, Ir joules per coulomb are transferred internally and V joules per coulomb remain for the external circuit.
- V
- terminal potential difference (V)
- ε
- emf (V)
- I
- current delivered (A)
- r
- internal resistance (Ω)
The term Ir is often called the lost volts. The energy is not destroyed; it normally appears as thermal energy inside the cell.
Complete circuit equation
If external resistance R is connected to the cell, the terminal voltage is V=IR. Combining this with V=ε−Ir gives:
The external and internal resistances are in series with respect to the circuit current, so both restrict the current supplied by the cell.
Open-circuit condition
A voltmeter of very high resistance draws negligible current. With I≈0, the internal drop Ir≈0, so the measured terminal voltage is approximately the emf.
What happens as load current increases
Reducing external resistance increases current. The internal drop Ir then becomes larger and terminal voltage falls. This explains why a battery's measured voltage can sag when it powers a heavy load.
When to use: For a source delivering current.
| Change | Result for a discharging cell |
|---|---|
| I increases | Ir increases and V decreases |
| r increases | Internal loss increases and V decreases |
| Circuit opens | I becomes zero and V becomes ε |
| R becomes very large | Current approaches zero and V approaches ε |
Short-circuit limit
If the terminals are joined by nearly zero external resistance, R≈0. Current is then limited mainly by the internal resistance.
This current can be very large, rapidly heating and damaging the cell and connecting wires.
Cell being charged
When an external supply forces current into the positive terminal of a cell, energy must overcome both the cell's emf and its internal resistance. Under this charging convention, the applied terminal voltage is:
When to use: For a cell being charged, with current entering its positive terminal.
Thus a charging source must provide a terminal voltage greater than the cell's emf.
Worked example
A cell has emf 12 V and internal resistance 0.50 Ω. It supplies 2.0 A.
The source supplies 12 J per coulomb: 11 J per coulomb reaches the external circuit and 1 J per coulomb is transferred internally.
Finding internal resistance experimentally
Measure terminal voltage V for several delivered currents I. Rearranging gives a linear relation:
A graph of V vertically against I horizontally has vertical intercept ε and slope −r.
Exam-ready reasoning
- Draw a real cell as emf ε in series with internal resistance r.
- For discharge, write ε=V+Ir and hence V=ε−Ir.
- Use V=IR for the external resistor.
- Check the open-circuit limit I=0 gives V=ε.
- Distinguish discharging V=ε−Ir from charging V=ε+Ir.
- Explain every voltage term as energy transferred per unit charge.
Exam Focus
Exam Focus
Important points
- Terminal p.d. meaning
- Real-source model
- V=ε−Ir
- Lost volts
- Complete-circuit current
- Open circuit
- Charging equation
- V-I graph
Important formulas
- V=ε-Ir
- I=εR+r
- V=ε RR+r
- V_{\mathrm{charge}}=\varepsilon+Ir
Common exam traps
- Choose the sign from whether the cell delivers or receives current.
- Open circuit means negligible current.
Long question structure
Real cell model → energy balance → discharge equation → complete circuit → limiting cases → charging → graph
Common Mistakes
Common Mistakes
- Mistake
Treating terminal voltage and emf as always equal.
- Mistake
Using V=ε+Ir while the cell is discharging.
- Mistake
Ignoring internal resistance in the complete circuit.
- Mistake
Calling internal heating destroyed energy.
- Mistake
Reading the V-I graph slope as +r instead of −r.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
A real source is modeled by emf ε in series with internal resistance r.
For a discharging cell, V=ε−Ir.
Ir is the internal voltage drop or lost volts.
With external resistance R, I=εR+r.
Terminal potential difference is measured across a source's terminals.
V=ε-Ir, I=εR+r
- Definition
- Terminal potential difference is measured across a source's terminals.
- Formula
- V=ε-Ir, I=εR+r
- Example
- For ε=12 V, r=0.50 Ω and I=2.0 A, V=11 V.
- Critical warning
- Use minus Ir for discharge and plus Ir for charging.
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Exam Practice
Definition
Q1. What is terminal potential difference?
Derivation
Q2. Derive V=ε−Ir for a discharging cell.
Reasoning
Q3. Why does terminal voltage fall when load current increases?
Numerical
Q4. A 12 V cell with r=0.50 Ω supplies 2.0 A. Find terminal voltage.
Graph
Q5. How can emf and internal resistance be found from a V-I graph?