A real cell loses some of its energy inside itself. The relationship E = V + Ir accounts for that, where E is the e.m.f., V the terminal voltage, I the current and r the internal resistance.
This lesson is part of SPM Physics electricity. It uses the series rules from solving series and parallel circuits.
How does the lost-volts picture work?
Think of the cell as an ideal source of e.m.f. E with a small resistor r in series. The current flows through r as well as the outside resistor R.
The voltage across r is Ir, the lost volts. The rest, V = E − Ir, is what reaches the external circuit.
Worked example: a cell driving a resistor
A cell of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor.
- Total resistance = R + r = 2.5 + 0.50 = 3.0 Ω.
- Current I = E ÷ (R + r) = 1.5 ÷ 3.0 = 0.50 A.
- Terminal voltage V = IR = 0.50 × 2.5 = 1.25 V.
- Lost volts Ir = 0.50 × 0.50 = 0.25 V.
Check: V + Ir = 1.25 + 0.25 = 1.50 V, which equals E.
Worked example: reading a graph
A student records two points for a cell: V = 1.40 V at I = 0.20 A, and V = 1.20 V at I = 0.60 A.
The gradient is (1.20 − 1.40) ÷ (0.60 − 0.20) = −0.20 ÷ 0.40 = −0.50 V A⁻¹, so r = 0.50 Ω. E is the intercept at I = 0: E = 1.40 + (0.20 × 0.50) = 1.50 V.
The mistake that loses marks
A common slip is to treat E = 1.5 V as the voltage across the resistor, which gives a current of 1.5 ÷ 2.5 = 0.60 A. The resistor only receives the terminal voltage, and the internal resistance also limits the current.
The correct current is 0.50 A. Always add r to the circuit resistance before applying Ohm’s law to the whole circuit.
Check yourself
A cell has e.m.f. 6.0 V and internal resistance 1.0 Ω. It is connected to a 5.0 Ω resistor. Find the current and the terminal voltage.
Answer
I = E ÷ (R + r) = 6.0 ÷ 6.0 = 1.0 A.
V = IR = 1.0 × 5.0 = 5.0 V. Lost volts Ir = 1.0 V, and 5.0 + 1.0 = 6.0 V, which matches E.
What to study next
Test the whole topic in the electricity practice set. The units and significant figure checker helps you check the unit of the gradient.
For more graph questions, see electrical reasoning from diagrams and measurements. For a teacher to go through your graphs with you, see online one-to-one Physics tuition or the one-hour trial class (from RM50).