understand the effects of the internal resistance of a source of e.m.f. on the terminal potential difference

Internal Resistance of a Source of e.m.f. – Effect on the Terminal Potential Difference

Learning Objectives

• Distinguish clearly between e.m.f. (ℰ) and terminal potential difference (V).
• Derive and apply the relation V = ℰ – I r for a source that possesses an internal resistance r.
• State and use Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL) in circuits that contain internal resistance.
• Determine experimentally the e.m.f. and internal resistance of a cell using the V–I method and the potentiometer (null) method.
• Apply series‑parallel combinations and the potential‑divider rule to practical circuits.
• Identify common sources of experimental error and describe how to minimise them.

Relevant Cambridge A‑Level Syllabus (9702)

Circuit Symbols Used in Practical Circuits (Cambridge Standard)

Symbol (schematic)	Meaning
	Ideal source of e.m.f. ℰ (no internal resistance)
	Cell containing internal resistance r (zig‑zag resistor drawn inside the cell symbol)
	Ammeter – placed in series, low resistance
	Voltmeter – placed in parallel, high resistance
	Potentiometer – uniform resistance wire with a sliding contact (jockey)
	Connecting wires – assumed ideal (negligible resistance)

Key Definitions (formal)

• e.m.f. (ℰ) – The maximum potential difference a source can maintain when no current flows (open‑circuit voltage).
• Internal resistance (r) – The inherent resistance of the source that causes a voltage drop Ir when a current  flows.
• Terminal potential difference (V) – The potential difference actually measured across the external terminals while the source supplies a current .
• Load resistance (R) – Any external resistor (or combination) connected to the source.
• Potential divider – Two (or more) resistors in series across a source; the voltage across each resistor is proportional to its resistance.

Kirchhoff’s Laws (explicit statements)

Theoretical Background – Derivation of V = ℰ – I r

1. Consider a cell of e.m.f. ℰ and internal resistance r supplying a current I to an external load R.
2. Applying KVL round the outer loop (cell + r + R):
   

   ℰ – I r – I R = 0

3. Re‑arranging gives the terminal potential difference across the load:
   

   V ≡ I R = ℰ – I r

4. Substituting Ohm’s law for the external circuit (I = V/R) leads to the linear relationship
   

   V = ℰ – r I

Consequences:

• When I = 0 (open circuit) → V = ℰ (the e.m.f.).
• When the cell is short‑circuited (R = 0) → I = ℰ/r (maximum current).
• Plotting V (y‑axis) against I (x‑axis) yields a straight line:

  • Slope = –r (negative because V falls as I increases).
  • Y‑intercept = ℰ.

Series and Parallel Combinations (recap)

• Series: \(R{\text{eq}} = R1 + R_2 + \dots\)
• Parallel: \(\displaystyle\frac{1}{R{\text{eq}}}= \frac{1}{R1}+ \frac{1}{R_2}+ \dots\)
• When a cell with internal resistance r is placed in series with external resistors, the total series resistance is r + R_eq.

Application of Kirchhoff’s Laws to a Circuit Containing Internal Resistance

Let the currents in the two parallel branches be \(I1\) (through \(R1\)) and \(I2\) (through \(R2\)). The total current supplied by the cell is \(I = I1 + I2\).

1. KCL at the junction: \(I = I1 + I2\)
2. KVL round the outer loop (cell → r → R₁ → back): \(\displaystyle ℰ - I r - I1 R1 = 0\)
3. KVL round the lower loop (cell → r → R₂ → back): \(\displaystyle ℰ - I r - I2 R2 = 0\)

Solving the three equations simultaneously gives:

\[

I1 = \frac{ℰ - I r}{R1},\qquad

I2 = \frac{ℰ - I r}{R2},\qquad

V = I1 R1 = I2 R2 = ℰ - I r

\]

Thus the terminal voltage is reduced by the internal drop \(I r\) irrespective of how the external resistors are arranged.

Potential Divider Rule

If two resistors \(Ra\) and \(Rb\) are in series across a source of e.m.f. ℰ (internal resistance ignored for the moment), the voltage across each resistor is

\[

Va = ℰ\frac{Ra}{Ra+Rb},\qquad

Vb = ℰ\frac{Rb}{Ra+Rb}

\]

This rule is used extensively in the potentiometer method (see below).

Potentiometer (Null‑Method) Principle

1. A uniform resistance wire of length \(L\) and total resistance \(Rp\) carries a steady current \(Ip\). The potential gradient along the wire is
   

   \[

   k = \frac{Ip Rp}{L}\;\;(\text{V m}^{-1})

   \]

2. A standard cell of known e.m.f. \(\mathcal{E}s\) is connected to the wire; the balance length \(\ells\) (where the galvanometer shows zero current) satisfies
   

   \[

   \mathcal{E}s = k\,\ells

   \]

3. Replacing the standard cell with the unknown cell (e.m.f. \(\mathcal{E}\)) and finding the new balance length \(\ell\) gives
   

   \[

   \mathcal{E} = k\,\ell

   \]

4. Eliminating \(k\) yields the simple ratio
   

   \[

   \boxed{\displaystyle \frac{\mathcal{E}}{\mathcal{E}s}= \frac{\ell}{\ells}\quad\Longrightarrow\quad \mathcal{E}= \mathcal{E}s\frac{\ell}{\ells}}

   \]

Because the galvanometer reads zero, no current is drawn from the cell under test, so the measured value is the true e.m.f., unaffected by internal resistance.

Experimental Determination of Internal Resistance – V–I Method

1. Connect the cell (with its internal resistance r) in series with an ammeter (A) and a variable load resistor R. Connect a voltmeter (V) across the load.
2. Choose a series of load resistances (e.g., 5 Ω, 10 Ω, 20 Ω, 40 Ω, 80 Ω). For each R record:

   • Current \(I\) from the ammeter.
   • Terminal voltage \(V\) from the voltmeter.

3. Enter the data in a table and plot \(V\) (vertical axis) against \(I\) (horizontal axis).
4. Fit a straight line (linear regression). From the equation \(V = ℰ - r I\):

   • Slope = \(-r\) → \(r = -(\text{slope})\).
   • Y‑intercept = ℰ → e.m.f. is read directly.

Sample Data (V–I Method)

Data Analysis Example

Using the five points above a linear regression gives

• Internal resistance: \(r = 0.53\;Ω\)
• e.m.f.: \(ℰ = 5.23\;V\)

Common Sources of Error & Mitigation Strategies

• Contact resistance at the cell terminals – Clean terminals, use tight spring clips, and repeat measurements to check consistency.
• Voltmeter loading – Ensure the voltmeter’s internal resistance is at least 10 × the largest load resistance; otherwise the measured voltage is reduced by the divider formed with the voltmeter.
• Ammeter resistance – Use a low‑resistance ammeter; if its resistance is not negligible, add it to the series resistance when analysing the data.
• Temperature rise of the cell – The internal resistance increases with temperature; allow the cell to cool between readings or record the temperature and correct if required.
• Reading and resolution errors – Use digital meters with appropriate range, record to the correct number of significant figures, and minimise parallax when reading analog scales.

Key Points to Remember

1. ℰ is the open‑circuit voltage (I = 0); V is the voltage actually delivered when current flows.
2. V falls linearly with I because of the internal drop \(I r\).
3. A single V–I graph provides both ℰ (y‑intercept) and r (negative slope).
4. The potentiometer measures ℰ without drawing current, giving the true e.m.f. regardless of r.
5. Series‑parallel resistor rules and Kirchhoff’s laws are essential for analysing any circuit that contains internal resistance.
6. Minimising instrument loading and contact resistance greatly improves the accuracy of the determined ℰ and r.

Practice Questions

1. Terminal voltage of a cell
   

   A cell has ℰ = 12.0 V and internal resistance r = 0.8 Ω. Find the terminal voltage when it supplies a current of 3 A.
   

   Solution: \(V = ℰ - I r = 12.0 - (3)(0.8) = 9.6\;V\).

2. Determine ℰ and r from V–I data
   

   Data: (I = 0.25 A, V = 9.8 V); (I = 0.40 A, V = 9.5 V); (I = 0.55 A, V = 9.2 V).
   

   Solution outline: Plot V vs I, calculate slope ≈ –0.6 Ω → r = 0.6 Ω; intercept ≈ 10.0 V → ℰ = 10.0 V.

3. Why must a voltmeter have a much larger resistance than the load?
   

   Answer: A voltmeter is connected in parallel with the load. If its resistance were comparable to the load, the two would form a voltage divider, reducing the voltage across the load and giving a measured V that is lower than the true terminal p.d.

4. Kirchhoff’s‑law calculation
   

   For the circuit in the “Kirchhoff’s Laws” figure, ℰ = 6 V, r = 0.5 Ω, \(R1=4 Ω\), \(R2=6 Ω\). Find the terminal voltage.
   

   Solution outline:

   • Parallel equivalent: \(R_{eq}= \frac{4\times6}{4+6}=2.4 Ω\).
   • Total series resistance: \(R{tot}= r + R{eq}=0.5+2.4=2.9 Ω\).
   • Current: \(I = ℰ / R_{tot}=6/2.9≈2.07 A\).
   • Terminal voltage: \(V = ℰ - I r = 6 - (2.07)(0.5)≈4.97 V\).

5. Potentiometer verification
   

   Describe how you would use a potentiometer to verify the e.m.f. obtained from the V–I method for the same cell.

Further Investigation Ideas

• State of charge: Measure r for a rechargeable battery at 100 %, 50 % and 10 % charge and comment on the trend.
• Temperature dependence: Gently heat the cell (e.g., with a lamp) and repeat the V–I experiment; plot r versus temperature.
• Aging effect: Compare the internal resistance of a brand‑new cell with that of a cell that has undergone several hundred charge‑discharge cycles.
• Potentiometer vs V–I method: Determine ℰ by both techniques for the same cell, calculate the percentage difference and discuss possible reasons (instrument loading, contact resistance, temperature, etc.).