Measuring E.m.f. & Internal Resistance | CAIE 9702 Paper 2 Notes

Paper 2

Physics9702

Measuring E.m.f. & Internal Resistance

One operating point cannot reveal both

E

and

r

. Change the external resistance to obtain several current and terminal-p.d. pairs.

The straight-line form of

V=E-Ir

then separates the two unknowns.

Collect paired readings

Change the variable resistor, and read V and I again. Each setting gives one pair of numbers, and every pair lies on the same straight line.

Connect the ammeter in series and the voltmeter across the source.
Change the variable resistor to change the current safely.
Record a terminal p.d. for each current.
Plot terminal p.d. $V$ vertically against current $I$ horizontally.

Read E and r from one line

Plot terminal p.d. against current and the battery gives you both numbers: the intercept is E, the gradient is minus r.

The vertical intercept is $E$ . At $I=0$ , no energy is transferred by the internal resistance, so $V=E$ .
The gradient is $-r$ . Internal resistance is the positive magnitude of that gradient.
If the line reaches $V=0$ , the current-axis intercept is the short-circuit current $E/r$ .

Your turn9702/11/M/J/25 Q37

A graph of terminal p.d. against current intercepts the p.d. axis at 1.6 V and the current axis at 80 mA. Calculate the internal resistance. [2] (9702/11/M/J/25 Q37)

Show worked answer

The gradient magnitude is $r=1.6/(80\times10^{-3})=20\ \Omega$ .