Physics (9630) | Electromotive force and internal resistance

Electromotive Force and Internal Resistance: Understanding Real Power Sources

Hello future Physicists! Welcome to a crucial topic in the "Electricity" section. Up until now, we've often treated power sources (like batteries or cells) as perfect providers of voltage. However, in the real world, batteries are a little bit complicated!

This chapter teaches us that every real power source contains a tiny hidden obstacle that reduces the voltage actually available to the circuit. Mastering this concept—Electromotive Force (emf) and Internal Resistance—is essential for accurate circuit analysis and practical applications. Let’s dive into what makes a battery tick (and sometimes run a little warm!).

What is Electromotive Force (emf), \(\mathcal{E}\)?

The Electromotive Force (\(\mathcal{E}\)) is the total energy supplied by a cell or battery per unit charge passing through it.

Emf vs. Potential Difference (p.d.)

It is vital not to confuse emf with standard potential difference (p.d.).

• Potential Difference (V): This is the work done per unit charge moving between two points outside the cell, usually across a component (like a resistor).
• Electromotive Force (\(\mathcal{E}\)): This is the total energy converted from chemical (or other) energy into electrical energy per unit charge, regardless of where that energy is used. It represents the "maximum potential" of the source.

Think of it like this: Imagine a water pump in a circuit.

• The Emf (\(\mathcal{E}\)) is the total force the pump is capable of generating to push the water (charge) through the whole system.
• The Potential Difference (V) is the energy used up only when the water hits an external obstacle (resistor).

The Emf Equation

The definition of emf is expressed mathematically as:

$$ \mathcal{E} = \frac{W}{Q} $$

Where:
\(\mathcal{E}\) is the Electromotive Force (measured in Volts, V)
\(W\) is the Work done (or energy supplied) by the source (measured in Joules, J)
\(Q\) is the Charge moved (measured in Coulombs, C)

Note: Since \(1 \, V = 1 \, J \, C^{-1}\), emf is measured in Volts.

Internal Resistance, \(r\): The Hidden Drain

When charge flows through a cell or battery, it has to move through the internal materials, such as the chemicals (electrolyte) and electrodes. These materials inevitably oppose the flow of charge, just like any external resistor.

Defining Internal Resistance

Internal resistance (\(r\)) is the resistance to the flow of charge provided by the components within the cell or power source itself.

• It is measured in Ohms (\(\Omega\)).
• Because of internal resistance, some electrical energy is converted into thermal energy (heat) inside the power source. This is why batteries get warm when used heavily!

Introducing "Lost Volts"

The energy wasted internally manifests as a drop in potential difference. This voltage drop across the internal resistance is often called the lost volts (\(V_{lost}\)).

Using Ohm’s law, the lost volts is calculated by:

$$ V_{lost} = I r $$

Where \(I\) is the current flowing through the circuit, and \(r\) is the internal resistance.

Analogy Check: If the pump (emf) generates a total pushing power of 12V, but 1V is lost immediately due to friction inside the pump itself (internal resistance), then only 11V is available to push water through the external pipes.

Did you know? Old batteries often have higher internal resistance because the chemicals inside are depleted or degraded, leading to higher lost volts and less power output.

The Electromotive Force (Emf) Circuit Equation

We can now relate the total energy supplied by the cell (Emf) to the energy used up both externally (across the load resistor $R$) and internally (across $r$). This is essentially an application of the conservation of energy (Kirchhoff’s Second Law).

The Full Equation

The total energy supplied per unit charge (\(\mathcal{E}\)) must equal the energy used per unit charge outside the cell (\(V\)) plus the energy used per unit charge inside the cell (\(V_{lost}\)).

$$ \mathcal{E} = V + V_{lost} $$

Substituting \(V_{lost} = Ir\), we get:

$$ \mathcal{E} = V + I r \quad \text{(Equation 1)} $$

Where:
\(\mathcal{E}\) = Electromotive force (V)
\(V\) = Terminal Potential Difference (V)
\(I\) = Current (A)
\(r\) = Internal resistance (\(\Omega\))

Expressing V using External Resistance (R)

The terminal p.d., \(V\), is simply the voltage across the external load resistor, \(R\). According to Ohm's Law, \(V = I R\).

Substitute \(V = I R\) into Equation 1:

$$ \mathcal{E} = I R + I r $$

Factor out \(I\):

$$ \mathcal{E} = I (R + r) \quad \text{(Equation 2)} $$

This equation shows that the total resistance in the circuit is the sum of the external load resistance (\(R\)) and the internal resistance (\(r\)).

Terminal Potential Difference (V): What the Circuit Sees

Definition of Terminal Potential Difference (V)

The terminal potential difference (V) is the actual voltage measured across the terminals of the power source when current is being drawn (i.e., when the circuit is closed). This is the voltage available to the external circuit components.

The Relationship between V and I

The terminal p.d. is highly dependent on the current flowing, $I$.

$$ V = \mathcal{E} - I r $$

• Open Circuit (Switch Open, \(I = 0\)): If no current is flowing, the lost volts (\(Ir\)) is zero. Therefore, \(V = \mathcal{E}\). This is how you measure the true emf of a cell.
• Closed Circuit (Current Flowing, \(I > 0\)): As current increases, the lost volts (\(Ir\)) increases. Consequently, the terminal p.d. (\(V\)) must decrease.
• Short Circuit (External Resistance \(R \approx 0\)): This draws maximum current, \(I_{max} = \mathcal{E}/r\). The terminal p.d. \(V\) drops close to zero, and the maximum power is dissipated internally (hence dangerous overheating).

Example: A 12V car battery (emf = 12V) might have an internal resistance of \(0.01 \, \Omega\). When the starter motor is activated, it draws 300 A.

Lost Volts: \(V_{lost} = I r = 300 \, A \times 0.01 \, \Omega = 3 \, V\).
Terminal P.D.: \(V = \mathcal{E} - V_{lost} = 12 \, V - 3 \, V = 9 \, V\).

This drop from 12V to 9V is why your car headlights (connected to \(V\)) dim significantly when you start the engine!

Required Practical 3: Determining \(\mathcal{E}\) and \(r\) Graphically

The syllabus requires you to understand how to investigate the emf and internal resistance of a cell using practical measurements. This relies on the equation \(V = \mathcal{E} - I r\).

Step-by-Step Graphical Method

We want to arrange the equation \(V = \mathcal{E} - I r\) into the form of a straight line graph: \(y = m x + c\).

Rearranging the equation:

$$ V = (-r) I + \mathcal{E} $$

1. **Set up the circuit:** Use an ammeter to measure current \(I\) and a voltmeter to measure the terminal p.d. \(V\) across the cell terminals. Include a variable resistor (or several fixed resistors) to change the external load \(R\).

2. **Collect data:** Measure pairs of \(V\) and \(I\) values for several different loads (i.e., different settings of the variable resistor).

3. **Plot the graph:** Plot Terminal Potential Difference, \(V\) (on the y-axis), against Current, \(I\) (on the x-axis).

4. **Analysis:** The resulting graph will be a straight line that slopes downwards.

• The y-intercept (where \(I=0\)) gives you the value of the emf, \(\mathcal{E}\).
• The gradient (slope, \(m\)) of the line is equal to \(-r\). Therefore, the magnitude of the gradient gives you the value of the internal resistance, \(r\).

This practical technique is a common exam question, so make sure you are confident linking \(V = (-r) I + \mathcal{E}\) to the graph coordinates!

Key Takeaway

Electromotive force (\(\mathcal{E}\)) is the total energy supplied by the source. Internal resistance (\(r\)) causes a voltage drop (lost volts, \(Ir\)) inside the source. The voltage available to the external circuit (terminal p.d., \(V\)) is always less than the emf when current flows, following the relationship: \(\mathcal{E} = V + Ir\).