The emf, internal resistance and the closed circuit
Week 2 · Chapter 1 (B): the emf of a cell and its internal resistance, Ohm’s law for a closed circuit I = VB/(R + r), terminal voltage V = VB − Ir and the emf as the open-circuit voltage, worked circuits (series, parallel, series-parallel, internal resistance) · 3 lessons
2
Ohm’s law for a closed circuit and terminal voltage
Physics Sec 3 Term 1, Chapter 1: pp. 13–17
GoalUse I = VB ÷ (R + r) and V = VB − Ir, and say why the emf is the voltage of a cell on open circuit.
New words
closed circuit · a complete circuit in which a current flowsterminal voltage (V) · the potential difference between the two poles of the source while it supplies a current: V = VB − Iropen circuit · a circuit with a break or an open switch, so that no current flows
Explanation
Putting the two parts together, VB = IR + Ir = I(R + r), so the current in a closed circuit is I = VB ÷ (R + r): the emf divided by the total resistance, outside and inside. The voltage between the two poles of the source while it supplies the current is the terminal voltage V. It is the voltage across the outside resistor, V = IR, and it is also V = VB − Ir, the emf minus the lost volts. When the current grows, Ir grows and the terminal voltage falls. On open circuit I = 0, so nothing is lost and V = VB: this is why the emf is the voltage of the cell with nothing connected, read by a voltmeter of very high resistance. A graph of V against I is a straight line that starts at VB when I = 0 and falls by r volts for every ampere, so its slope is −r. Two readings are enough to find both numbers: first r from the fall in voltage divided by the rise in current, then VB.
Worked examples
VB = 12 V, r = 0.5 Ω and R = 5.5 Ω. I = 12 ÷ (5.5 + 0.5) = 2 A. The terminal voltage is V = VB − Ir = 12 − 2 × 0.5 = 11 V, and the check IR = 2 × 5.5 = 11 V agrees.
VB = 9 V, r = 1.5 Ω and R = 4.5 Ω. I = 9 ÷ (4.5 + 1.5) = 1.5 A. V = 9 − 1.5 × 1.5 = 6.75 V, and IR = 1.5 × 4.5 = 6.75 V.
Reading
I (A)
V (V)
first
1
11.5
second
3
10.5
change
+2
−1.0
Invented readings of a cell: V = 11.5 V at I = 1 A and V = 10.5 V at I = 3 A. The voltage falls by 1.0 V while the current rises by 2 A, so r = 1.0 ÷ 2 = 0.5 Ω. Then VB = V + Ir = 11.5 + 1 × 0.5 = 12 V.
Class activity
On graph paper, plot the two invented readings of Example 3 (current on the horizontal axis, voltage on the vertical axis), draw the straight line through them and extend it back to I = 0. Read VB where the line meets the voltage axis, and check that the fall in volts per ampere equals r. No batteries are needed.
Practice
1
VB = 24 V, r = 2 Ω and R = 10 Ω. Find the current and the terminal voltage.
I = 24 ÷ (10 + 2) = 2 A. V = 24 − 2 × 2 = 20 V, and IR = 2 × 10 = 20 V.
2
A cell with VB = 3 V and r = 0.6 Ω delivers 2 A. Find its terminal voltage.
V = VB − Ir = 3 − 2 × 0.6 = 1.8 V.
3
A cell reads 6.0 V on open circuit and 5.1 V while it delivers 1.5 A. Find its internal resistance.
Ir = 6.0 − 5.1 = 0.9 V, so r = 0.9 ÷ 1.5 = 0.6 Ω.
4
Explain why a torch bulb gets dimmer as its cells age.
The internal resistance of old cells grows, so the current through the bulb falls and more of the emf is lost inside the cells; the voltage across the bulb drops.
