Recall and use the equation for p.d. V = W / Q

4.2.3 Electromotive Force (EMF) and Potential Difference

Learning objectives (AO1)

• Recall the definitions of electromotive force (EMF) and potential difference (p.d.).
• Write and use the three core equations:

  • EMF \( \displaystyle \varepsilon = \frac{W}{Q}\)
  • Potential difference \( \displaystyle V = \frac{W}{Q}\)
  • Terminal p.d. with internal resistance \( \displaystyle V = \varepsilon - I r\)

• Explain how a voltmeter (analogue or digital) is used to measure p.d. and EMF, and how to choose an appropriate range.
• Describe practical methods for measuring EMF and p.d. (simple circuit and potentiometer).
• Interpret a V‑I graph for a source with internal resistance and determine \(r\) by extrapolation (AO2).

Key definitions (syllabus wording)

Units summary

Fundamental equations

1. EMF \( \displaystyle \varepsilon = \frac{W}{Q}\)
2. Potential difference \( \displaystyle V = \frac{W}{Q}\)
3. Terminal p.d. with internal resistance \( \displaystyle V = \varepsilon - I r\)

Derivation of \(V = \dfrac{W}{Q}\)

1. By definition, the work required to move a charge \(Q\) through a potential difference \(V\) is \(W = V Q\).
2. Re‑arrange for \(V\): \(V = \dfrac{W}{Q}\).
3. The same reasoning applied to a source moving the charge round a *complete* circuit gives the EMF formula \(\varepsilon = \dfrac{W}{Q}\).

Important relationships

• When the circuit is open (\(I = 0\)), no current flows through the internal resistance, so the terminal p.d. equals the EMF:

  \[

  V_{\text{open}} = \varepsilon .

  \]

• For a real source the internal resistance causes a drop \(I r\) in the direction of current flow. Consequently,

  \[

  V = \varepsilon - I r\quad\text{(current leaves the positive terminal)}.

  \]

• Plotting measured terminal p.d. (\(V\)) against current (\(I\)) gives a straight line.

  • The V‑axis intercept is the EMF \(\varepsilon\).
  • The negative slope equals the internal resistance \(-r\).

  This graphical method is a common AO2 requirement.

Using a voltmeter

• Connect the voltmeter in parallel with the component whose p.d. is to be measured.
• Analogue voltmeter:

  • Scale is marked in volts; the needle points to the reading.
  • Choose a range that is just above the expected p.d. (e.g. 0–12 V for a 9 V cell). This gives the best resolution and accuracy.

• Digital voltmeter:

  • Displays a numeric value; many have auto‑range, otherwise select a range manually.
  • Resolution is typically 0.01 V or 0.001 V depending on the model.

• Because the voltmeter’s internal resistance is very high, it draws only a negligible current and does not significantly alter the circuit.

Practical methods for measuring EMF and p.d.

Method 1 – Simple series circuit (no potentiometer)

1. Connect the source (cell) in series with a known external resistor \(R\).
2. Place an ammeter in series to measure the current \(I\).
3. Connect a voltmeter across the external resistor (parallel to \(R\)) to read the terminal p.d. \(V\).
4. If the internal resistance \(r\) is known, calculate the EMF:

   \[

   \varepsilon = V + I r .

   \]

5. If \(r\) is unknown, repeat the experiment with a second value of \(R\) (different current). Plot \(V\) against \(I\); the intercept gives \(\varepsilon\) and the slope gives \(-r\).

Method 2 – Potentiometer (advanced, optional)

1. Set up a long uniform wire with a known current; the potential gradient \(\frac{dV}{dx}\) along the wire is constant.
2. Connect the unknown source to a sliding contact on the wire and balance a galvanometer so that no current flows through the source.
3. The length \(l\) of wire required to balance the source gives the EMF directly:

   \[

   \varepsilon = \left(\frac{dV}{dx}\right) l .

   \]

4. This method measures EMF without drawing current from the source, so the reading equals \(\varepsilon\) exactly.

Determining internal resistance from measurements (AO2)

Two common approaches:

1. Graphical extrapolation – Measure \(V\) for at least two different loads, plot \(V\) (y‑axis) against \(I\) (x‑axis). The straight‑line fit gives:

   • Intercept on the V‑axis = \(\varepsilon\).
   • Slope = \(-r\) → \(r = -\text{slope}\).

2. Two‑load calculation – Using the equation \(V = \varepsilon - I r\) for two different loads:

   \[

   \begin{aligned}

   V1 &= \varepsilon - I1 r,\\

   V2 &= \varepsilon - I2 r.

   \end{aligned}

   \]

   Subtract the equations to eliminate \(\varepsilon\):

   \[

   r = \frac{V1 - V2}{I2 - I1}.

   \]

   Then substitute back to find \(\varepsilon\). This method requires only two sets of readings.

Example calculations

Example 1 – EMF from work

A source does 18 J of work to move 3 C of charge once round a complete circuit. Find its EMF.

\[

\varepsilon = \frac{W}{Q} = \frac{18\ \text{J}}{3\ \text{C}} = 6\ \text{V}

\]

Example 2 – Terminal p.d. with internal resistance

A 12 V battery has an internal resistance of 0.5 Ω and supplies a current of 4 A. Find the terminal p.d.

\[

V = \varepsilon - I r = 12\ \text{V} - (4\ \text{A})(0.5\ \Omega) = 12\ \text{V} - 2\ \text{V} = 10\ \text{V}

\]

Example 3 – Determining internal resistance (two‑load method)

Using a 9 V cell, the following readings are obtained:

Calculate \(r\) and \(\varepsilon\).

\[

r = \frac{V1 - V2}{I2 - I1}

= \frac{7.5 - 8.1}{0.9 - 1.5}

= \frac{-0.6}{-0.6}

= 1\ \Omega .

\]

\[

\varepsilon = V1 + I1 r = 7.5\ \text{V} + (1.5\ \text{A})(1\ \Omega) = 9.0\ \text{V}.

\]

Common misconceptions

• “EMF always equals terminal p.d.” – True only for an ideal source (no internal resistance) or when the circuit is open (\(I=0\)).
• “\(V = W/Q\) only works for batteries.” – The relation applies to any situation where work is done moving charge, e.g. a discharging capacitor.
• Sign of work – If the source supplies energy, \(W\) (and therefore \(\varepsilon\) or \(V\)) is positive. If work is done *on* the source, the sign is negative.
• Direction of current – In the equation \(V = \varepsilon - I r\) the current is taken as flowing away from the positive terminal of the source. Reversing the current changes the sign of the \(I r\) term.

Practice questions

1. A 9 V battery supplies a current of 0.5 A through a circuit that has an internal resistance of 2 Ω. Calculate the terminal p.d.
2. What amount of work is required to move 0.02 C of charge through a potential difference of 15 V?
3. If a source has an EMF of 12 V and the terminal p.d. measured under load is 10 V, what is the current flowing when the internal resistance is 0.5 Ω?
4. Explain how you would use a digital voltmeter to measure the EMF of a cell, stating how you would choose the range.
5. In a circuit with a 6 Ω external resistor, a battery of EMF 9 V and internal resistance 1 Ω supplies a current of 1 A. Verify the terminal p.d. using the equation \(V = \varepsilon - I r\).
6. Two loads give the following readings for a cell:

   Load (Ω)	Current (A)	Terminal p.d. (V)
   5	1.2	7.8
   10	0.7	8.2

   Determine the internal resistance and EMF of the cell.

Answers to practice questions

1. \(V = \varepsilon - I r = 9\ \text{V} - (0.5\ \text{A})(2\ \Omega) = 8\ \text{V}\).
2. \(W = V Q = (15\ \text{V})(0.02\ \text{C}) = 0.30\ \text{J}\).
3. \(\varepsilon - V = I r \Rightarrow 12\ \text{V} - 10\ \text{V} = I(0.5\ \Omega)\) → \(2\ \text{V} = 0.5 I\) → \(I = 4\ \text{A}\).
4. Connect the digital voltmeter across the cell’s terminals (parallel). Select a range just above the expected EMF – for a 9 V cell choose the 0–12 V range (or let the auto‑range function select it). Because the voltmeter draws negligible current, the reading equals the EMF.
5. Voltage drop across internal resistance: \(I r = (1\ \text{A})(1\ \Omega) = 1\ \text{V}\).

   Terminal p.d.: \(V = \varepsilon - I r = 9\ \text{V} - 1\ \text{V} = 8\ \text{V}\).

   Check with the external resistor: \(V_{\text{ext}} = I R = (1\ \text{A})(6\ \Omega) = 6\ \text{V}\).

   The remaining 2 V is the drop across the internal resistance, confirming the calculation.

6. Using the two‑load method:

   \[

   r = \frac{V1 - V2}{I2 - I1}

   = \frac{7.8 - 8.2}{0.7 - 1.2}

   = \frac{-0.4}{-0.5}

   = 0.8\ \Omega .

   \]

   EMF: \(\varepsilon = V1 + I1 r = 7.8\ \text{V} + (1.2\ \text{A})(0.8\ \Omega) = 8.76\ \text{V}\) (≈ 8.8 V).

Suggested diagram

Summary

• Potential difference is the work done per unit charge: \(V = \dfrac{W}{Q}\).
• EMF is the same relation applied to a source completing a full circuit: \(\varepsilon = \dfrac{W}{Q}\).
• Real sources have internal resistance \(r\); when current flows the terminal p.d. is reduced:

  \[

  V = \varepsilon - I r .

  \]

• When the circuit is open (\(I = 0\)), \(V = \varepsilon\).
• Use a voltmeter in parallel, choose a range just above the expected voltage for best accuracy, and interpret V‑I data to find both EMF and internal resistance.
• These concepts and techniques satisfy all the core requirements of Cambridge IGCSE Physics 0625 for topic 4.2.3.