1. Emf and Internal Resistance To do Work in groups of 4 or 5 to carry out Activity 2 – Internal Resistance. Internal Resistance V In theory… V = 12 V In practice… V = 11.8 V Why??? Notes p.11

2. Reminder …. Voltage = Energy per Coulomb When current flows, wires/battery etc heat up hence energy is “lost” per coulomb. (So we have some “lost volts” before the charge even reaches the load resistor.) The e.m.f. is the energy supplied per coulomb BEFORE any volts are lost. To measure e.m.f. we must have no current flowing so that there is no energy loss. V battery R =

3. In the example above we “lost” 0.2 V. We say that the battery had an “internal resistance” which used up the 0.2 V. Our circuit can be illustrated as 1. The voltage (V) measured across the battery terminals = voltage across the load, R. It is called the terminal potential difference, t.p.d.. 100  V - terminal + terminal r e.m.f. = 12 V = 11.8 V R

4. 2.Inside the battery, effectively, we have a supply e.m.f. of 12 V and a small internal resistance, “r”. 3.To measure e.m.f. remove the load, R, leaving an air gap so no current flows. 100  12V r e.m.f. = 12 V So V = e.m.f. = 12 V

5. 4.To calculate “r”, replace the load resistor, R, then 1 st find I : R = 100  V = 11.8 V I = ? I = V/R = 11.8 / 100 = 0.118 A 2 nd find r : lost volts = 12 – 11.8 = 0.2 V I = 0.118 A r = ? r = lost volts I = 0.2 / 0.118 = 1.69 

6. Worked Examples (see handout) Example 1

7. Example 2 : The voltage across a cell was recorded for various values of current, using the following circuit. V r R A

8. A graph of the results was plotted: 0.25 0.50 0.75 1.00 0.4 0.8 1.2 1.6 Current (A) 0 Voltmeter reading (V)

9. 0.25 0.50 0.75 1.00 0.4 0.8 1.2 1.6 Current (A) 0 Voltmeter reading (V) a) Determine the e.m.f. of the cell. 2 V (extrapolate line back to when I = 0 A) b) Calculate the internal resistance of the cell. PICK A POINT AND WRITE DOWN ALL YOU CAN! emf = 2V lost volts = 2 – 0.4 = 1.6V I = 1 A r = ? r = lost volts / I = 1.6 / 1 = 1.6 

10. c) Determine the short circuit current for this circuit. This means all volts (e.m.f.) used in r! The battery’s “r” is the only resistance in the circuit. e.m.f. = 2 V I = ? r =  I short circuit = e.m.f. r = 2 1.6 = 1.25 A

11. Summary 1.E.m.f is measured when no current flows. e.m.f. = I R T = I (R + r) 2.t.p.d. is voltage across the load resistance, R: t.p.d. = e.m.f. – lost volts t.p.d. = IR Lost volts = I r 3. “lost volts” is the potential difference across the internal resistance of the cell. 4. Short circuit current … I short = emf / r

12. Summary (cont.) 5. Ideal Electrical Supplies: The maximum power is transferred from a battery when the internal resistance = load resistance

13. Over to you – continue problems up to p.13-18, Q. 6 - 18

14. Problems on Circuitspage 60-61, Q 11-24 11. a) 6  b) 3  c) 2 A for A1, 1.5A for A2 d) 6V e) 24W 12. a) 11.5A b) 12.8A (12  ), 9.6A (8  ), 3.2A (24  ) 13. a) R1 = 230 , R2 = 115  b) LOW = 230W, HIGH = 690W 14.a) 4  b) 36W 15.a) 2V b) 1.6V c) r = 0.5  R = 2  d) 1.3A, 1.3V 16. 10  17. 5  18. 12  19.a) 1.3V b) current decreases as total resistance increases. c) tpd increases - as R increases it takes a larger share of the voltage OR as current increases there are more “lost volts” in the internal resistance of the cell.

15. Problems on Circuitspage 60-61, Q 11-24 16. 10  17. 5  18. 12  19.a) 1.3V b) current decreases as total resistance increases. c) tpd increases - as R increases it takes a larger share of the voltage OR as current increases there are more “lost volts” in the internal resistance of the cell. 20. 0.3A 21. a) 3  b) I = 0.67A, R = 3.73  22. 4V 23. a) standard circuit (see notes) b) (i) 1.1V (ii) approx 4.10   c) 4.00 ,  4.00 ,  4.17 ,  4.00 ,  4.10 ,  4.17 . d) (4.10 ± 0.02 )  24. a) 6V b) 0.1  c) 60A d) (i) 5.63V (ii) 21.1W