E.m.f. and internal resistance flashcards

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E.m.f. (ε) of a source
The energy transferred to electrical energy per unit charge passing through the source. Unit V = J C⁻¹.
Potential difference (compared with e.m.f.)
The energy transferred from electrical energy per unit charge passing through a component. Same unit as e.m.f., opposite direction of transfer.
Internal resistance r
The resistance of the materials inside a source, which dissipates energy whenever current flows through the source.
Terminal p.d.
The p.d. across the terminals of a source when it delivers a current: V = ε − Ir.
Lost volts
The p.d. across the internal resistance, Ir: the energy per coulomb dissipated as thermal energy inside the source.
Short circuit
A connection of (almost) zero resistance across the terminals; the current is the maximum ε/r and the terminal p.d. is zero.
E.m.f., current and resistances
ε = I(R + r), with R the total external resistance. (No — learn it)
Terminal p.d. of a real source
V = ε − Ir (expand ε = I(R + r) and use V = IR). (No — derive it)
E.m.f. as energy per charge
ε = W/Q (same form as V = W/Q). (No — learn it)
Maximum (short-circuit) current
I_max = ε/r (put R = 0 in ε = I(R + r)). (No — derive it)
Power balance in a circuit with a real source
εI = VI + I²r: power supplied = power to the load + power wasted inside. (No — derive it)
Fraction of energy reaching the load
V/ε = R/(R + r). (No — derive it)
Cells in series
Total e.m.f. = sum of e.m.f.s (a reversed cell subtracts); total internal resistance = sum of all the r values. (No — learn it)
Graph of terminal p.d. V against current I for a cell
A straight line with negative gradient: V-intercept = ε, gradient = −r, I-intercept = ε/r.
Core Practical 8 circuit
Cell in series with an ammeter, a variable resistor and a switch; voltmeter directly across the cell's terminals.
Power delivered to a load against load resistance R
Rises from zero, peaks when R = r, then falls slowly; at the peak only 50% of the energy reaches the load.
Model of a real cell in a circuit diagram
A perfect e.m.f. ε in series with a resistor r, drawn inside a dashed box; only the two terminals are accessible.
Why does a high-resistance voltmeter alone across a cell read (almost) its e.m.f.?
It draws almost no current, so the lost volts Ir are almost zero and V ≈ ε.
Why does the terminal p.d. fall as more current is drawn?
The lost volts Ir rise with the current, and V = ε − Ir.
Why do a car's headlamps dim while the starter motor runs?
The large starter current gives large lost volts, so the terminal p.d. across the lamps falls.
Why is a high-voltage laboratory supply given a very large internal resistance?
It limits the current to a small, safe value (ε/r) even if the terminals are touched or shorted.
Core Practical 8: why open the switch between readings?
To stop the cell heating up and running down, which would change r (and ε) during the experiment.
Core Practical 8: what does a protective resistor in series do?
Limits the current if the variable resistor is set to zero; it does not affect V against I because the voltmeter is across the cell only.
Effect of a voltmeter zero error on the Core Practical 8 results
The line shifts up or down: ε (intercept) is wrong, r (gradient) is unchanged.
ε = 1.50 V, r = 0.80 Ω, R = 4.2 Ω. Current and terminal p.d.?
I = 1.50 ÷ 5.0 = 0.300 A; V = 0.300 × 4.2 = 1.26 V (lost volts 0.24 V).
A voltmeter reads 6.20 V across a battery alone and 5.40 V with a 5.0 Ω resistor connected. Internal resistance?
I = 5.40 ÷ 5.0 = 1.08 A; r = (6.20 − 5.40) ÷ 1.08 = 0.74 Ω.
Battery ε = 12.6 V, r = 0.020 Ω. Short-circuit current?
I_max = ε/r = 12.6 ÷ 0.020 = 630 A.
ε = 9.0 V, r = 3.0 Ω, R = 6.0 Ω. Power to the load and power wasted inside?
I = 1.0 A; load I²R = 6.0 W; inside I²r = 3.0 W (supplied εI = 9.0 W).
Four 1.5 V, 0.30 Ω cells in series, one the wrong way round. Total ε and r?
ε = 1.5 + 1.5 − 1.5 + 1.5 = 3.0 V; r = 4 × 0.30 = 1.20 Ω.
A V–I line has V-intercept 1.60 V and passes through (0.80 A, 1.20 V). ε and r?
ε = 1.60 V; gradient = −0.40 ÷ 0.80 = −0.50 V A⁻¹, so r = 0.50 Ω.
A battery gives 0.50 A in 4.0 Ω and 0.30 A in 8.0 Ω. ε and r?
0.50(4.0 + r) = 0.30(8.0 + r) → r = 2.0 Ω; ε = 0.50 × 6.0 = 3.0 V.
Using I = ε/R for a real cell
Wrong: the current is ε/(R + r). The internal resistance is part of the circuit.
A cell connected the wrong way round in series
Its e.m.f. subtracts (total falls by 2ε) but its internal resistance still adds.
Finding r from a V–I graph
r = −gradient, from a large triangle on the line. Never use V/I at one point — that is the external resistance.
Reading ε from the intercept of a V–I graph
Only valid if the I-axis starts at zero; otherwise use ε = V + Ir at one point on the line.
Maximum power transfer and efficiency
Power to the load is greatest when R = r, but then only 50% of the energy reaches the load — maximum power is not maximum efficiency.

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