2. Definition of Potential Difference p.d. of 1 Volt when 1 joule of work is required to move 1 coulomb of charge. Example Calculate the work done in moving 2 C of charge across a p.d. of 10 V. W = ?V = 10 VQ = 2C W = Q x V W = 2 x 10 = 20 J

3. Internal Resistance Terminal potential difference ( V t.p.d. ) is the p.d. across the load Vtpd

4. Internal Resistance 2  Resistance of power supply itself  Work is done to push charges through power supply hence ‘Lost Volts’ Electromotive Force (E.M.F. ) Maximum energy to push unit charge around circuit  Terminal potential difference, work to push unit charge through external circuit ( load )

5. Internal Resistance 3 Apply the principle of conservation of energy Electromotive force = terminal p.d. + lost volts E.M.F ( E ) = V tpd + V lost But V lost = I x r I = current flowing and r = internal resistance Therefore E = V tpd + Ir

6. Internal Resistance 4  E = V t.p.d. + V lost  V t.p.d. = E - V lost V t.p.d = E – (I x r) V t.p.d I E.M.F open circuit p.d. NO lost volts - slope = r Short circuit current E = V lost 0 V t.p.d V t.p.d = -r x I + E y = mx + c

7. Internal Resistance 5 EMF is measured by connecting a voltmeter across the battery when there is NO load. No load means NO current and hence NO lost volts. Short Circuit Current is the maximum current the battery can deliver. From the graph, V tpd = 0V when current is maximum. Hence E = V lost. This is a theoretical value.

8. Example A cell of e.m.f. 1.5V and internal resistance 0.75 Ω is connected as shown in the following circuit. (a) Calculate the value of the reading on the voltmeter. (b) What is the value of the “ lost volts ” in this circuit? (a)V e.m.f. = I(R+r) 1.5 = I (3 + 0.75) I = 1.5/3.75 = 0.4 A V t.p.d. = IR = 0.4 x 3 = 1.2 V (b)Lost volts = Ir = 0.4 x 0.75 = 0.3 V OR Lost volts = V e.m.f - V t.p.d. = 1.5 – 1.2 = 0.3 V 1.5V 3 Ω V 0.75 Ω Or use voltage divider equation Total resistance

9. Internal Resistance 7 Example Calculate : a)The short circuit current of the cell and b) The quantity of heat energy dissipated in the battery when connected as shown. E = 6.0 V r= 1.0 Ω 4.0 Ω a) a)When short circuit current flows V tpd = 0V Therefore E = V l = I r I = E / r = 6.0 / 1.0 = 6.0 A

10. Internal Resistance 7a b) Energy per second ( power ) can be calculated by various ways : P = I 2 r P = 1.2 2 x 1.0 P = 1.44 A I = E / R total I = E / ( R = r ) I = 6.0 / ( 4.0+1.0 ) I = 1.2 A This is one more significant figure than data. We are allowed to quote ‘2’extra sig figs

11. Power Supplies The ideal voltage source maintains a constant output voltage no matter what load is connected to it. This ideal source would have ZERO internal resistance. In the real world power supplies have internal resistance and as soon as a load is connected then the terminal potential difference decreases. It is important to carefully match the internal resistance of the power supply with the resistance of the load: R = 5Ω Internal resistance, r, Load, R. If r > R then the ‘lost volts ‘ will be bigger than the terminal potential difference and lots of heat is generated inside supply

12. Amplifiers need to be matched carefully to loudspeakers of certain impedance if max power is to be transferred to the speakers. The internal resistance of the amplifier should equal that of the speakers for max power transfer.