The electromotive force, written as \( \mathcal{E} \), is the total energy supplied by a power source (like a battery) to move one coulomb of charge around a circuit.
It is also the maximum voltage a power source can provide when no current is flowing.
Formula: \( \mathcal{E} = I(R + r) \)
where:
Every real battery or power source has some resistance inside it. This is called internal resistance, denoted as \( r \).
It reduces the energy available to the rest of the circuit when current flows.
The larger the current, the more energy is "lost" inside the battery due to this resistance.
Used in Formula: \( \mathcal{E} = V + Ir \)
where \( Ir \) is the voltage lost due to internal resistance.
Lost voltage is the portion of the EMF that is used up inside the battery to overcome its internal resistance.
It’s also called the voltage drop across the internal resistance.
\( \text{Lost Voltage, v} = Ir \)
This value increases when the current increases or the internal resistance increases.
The terminal potential difference \( V \) is the voltage that is actually available at the terminals of the battery
for use in the external circuit. It is always less than the EMF when current is flowing, due to the lost voltage.
\( V = \mathcal{E} - Ir \)
\( V = IR \)
When no current flows, \( V = \mathcal{E} \).
Example 1: A battery with an electromotive force (EMF) of 12V is connected in series with a 4Ω resistor. The current flowing through the circuit is measured to be 2A. Calculate: (i) the internal resistance of the battery, and (ii) the terminal potential difference of the battery.
Example 2: A battery is rated 24V and has an internal resistance of 1.5Ω. It is connected to an external load of 10.5Ω. Calculate: (i) the current flowing in the circuit, and (ii) the voltage drop across the internal resistance (lost voltage).
Example 3: A battery of electromotive force (e.m.f.) 10V and internal resistance of 2Ω is connected to an external resistor of 6Ω. Calculate the potential difference (p.d.) across the terminals of the battery.
Total resistance in the circuit:
$$ R_{\text{total}} = R + r = 6 + 2 = 8Ω $$Current in the circuit:
$$ I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{10}{8} = 1.25A $$Now, use the formula for terminal potential difference:
$$ V = Ir $$ $$ V = (1.25 \times 6) $$ $$ = 7.5V $$Example 4: Two identical cells, each with an e.m.f. of 2V and internal resistance of 1.0Ω, are connected in parallel. This parallel combination is connected to an external load of 1.5Ω. Calculate the total current flowing in the circuit.
Since the cells are identical and connected in parallel, the total e.m.f. remains the same as that of a single cell.
$$ \mathcal{E}_{\text{total}} = 2\,\text{V} $$The internal resistances are in parallel, so we calculate the equivalent internal resistance using the formula for resistors in parallel:
$$ \frac{1}{r_{\text{eq}}} = \frac{1}{1} + \frac{1}{1} = \frac{2}{1} $$ $$ r_{\text{eq}} = \frac{1}{2} = 0.5\,\Omega $$The total resistance in the circuit is the sum of the equivalent internal resistance and the external load resistance:
$$ R_{\text{total}} = r_{\text{eq}} + R = 0.5 + 1.5 = 2\,\Omega $$Using Ohm's law, the total current in the circuit is:
$$ I = \frac{\mathcal{E}_{\text{total}}}{R_{\text{total}}} = \frac{2}{2} = 1\,\text{A} $$Example 6: A resistor of resistance \( R \) is connected across a cell. If the terminal potential difference of the cell is reduced to one-quarter of its e.m.f., express the internal resistance \( r \) of the cell in terms of \( R \). (NECO)
From the relationship between e.m.f. \( \mathcal{E} \), terminal voltage \( V \), current \( I \), external resistance \( R \), and internal resistance \( r \), we use:
$$ \mathcal{E} = V + Ir $$Also, since the external resistor \( R \) is connected across the cell, by Ohm’s law:
$$ I = \frac{V}{R} $$We are told that the terminal voltage is one-quarter of the e.m.f., so:
$$ V = \frac{1}{4} \mathcal{E} $$Now substitute \( I = \frac{V}{R} \) into the e.m.f. equation:
$$ \mathcal{E} = V + \left( \frac{V}{R} \right) r $$Factor out \( V \) on the right-hand side:
$$ \mathcal{E} = V \left(1 + \frac{r}{R} \right) $$Now substitute \( V = \frac{1}{4} \mathcal{E} \) into the equation:
$$ \mathcal{E} = \frac{1}{4} \mathcal{E} \left(1 + \frac{r}{R} \right) $$Divide both sides by \( \mathcal{E} \) (assuming \( \mathcal{E} \ne 0 \)):
$$ 1 = \frac{1}{4} \left(1 + \frac{r}{R} \right) $$Multiply both sides by 4:
$$ 4 = 1 + \frac{r}{R} $$Subtract 1 from both sides:
$$ 3 = \frac{r}{R} $$Multiply both sides by \( R \):
$$ r = 3R $$Therefore, the internal resistance of the cell is three times the external resistance: \( r = 3R \).
Example 5: An electric bell takes a current of 0.2A from a battery of two dry cells connected in series. Each cell has an e.m.f. of 1.5V and an internal resistance of 1.0Ω.
(i) Calculate the effective resistance of the bell.
(ii) What current would the bell take if the cells were arranged in parallel?
(i) When cells are connected in series:
Total e.m.f. of the battery is:
$$ \mathcal{E}_{\text{series}} = 1.5 + 1.5 = 3.0\,\text{V} $$Total internal resistance in series:
$$ r_{\text{series}} = 1.0 + 1.0 = 2.0\,\Omega $$Let \( R \) be the resistance of the bell. Using Ohm’s law:
$$ I = \frac{\mathcal{E}}{R + r} $$Substituting values:
$$ 0.2 = \frac{3}{R + 2} $$Multiply both sides by \( R + 2 \):
$$ 0.2(R + 2) = 3 $$ $$ R + 2 = \frac{3}{0.2} = 15 $$ $$ R = 15 - 2 = 13\,\Omega $$Therefore, the resistance of the bell is 13Ω
(ii) When the cells are connected in parallel:
For identical cells in parallel, the total e.m.f. remains the same as that of a single cell:
$$ \mathcal{E}_{\text{parallel}} = 1.5\,\text{V} $$The equivalent internal resistance for two identical resistors in parallel is:
$$ \frac{1}{r_{\text{parallel}}} = \frac{1}{1.0} + \frac{1}{1.0} = 2 $$ $$ r_{\text{parallel}} = \frac{1}{2} = 0.5\,\Omega $$Now calculate the total current using Ohm's law:
$$ I = \frac{\mathcal{E}}{R + r} = \frac{1.5}{13 + 0.5} $$ $$= \frac{1.5}{13.5} \approx 0.111\,\text{A} $$Therefore, the current the bell would take in parallel is approximately 0.111A.