1. Current, Voltage measurements; Potentiometers and bridges Explain how to extend the range of an ammeter Explain how to extend the range of an voltmeter Describe a potentiometer Explain how an emf can be measured exactly using a driver cell Outline the Wheatstone Bridge

2. Current can be measured using a moving coil ammeter or a digital multi-meter, which is put in series with the circuit. We have a circuit and ammeter. What should we do? Cut the circuit and reconnect it so that the ammeter is in series: Current and voltage measurements Has a known FSD (Full Scale Deflection)

3. Current Measurement What can we do if current is too high: the full scale deflection of the meter is not enough to measure the current? Analogy: to protect us against flood we can dig out a channel to redirect extra water! So we should attach a new wire to allow current flow along this new path (not through our ammeter). The range of an ammeter may be extended by inserting a shunt resistance in parallel.

4. Current Measurement The range of a meter may be extended by inserting a shunt resistance in parallel. What should we do if the current is too high? Example: if we remove one bulb from the circuit, then the current will be too high to measure. Indeed the current with two bulbs is already close to the full scale deflection of our meter. This ammeter cannot measure current of a circuit with one bulb only. ?

5. Current Measurement The range of a meter may be extended by inserting a shunt resistance in parallel. Some current flows through the resistance which is in parallel with the ammeter. Thus the ammeter can measure current which is higher than full scale deflection of the meter. This is because the actual current I is split into two currents, I 1 and I 2. I I1I1 I2I2

6. Current Measurement How to calculate what resistance to use? Applying Kirchoff’s current law for node A: I=I 1 +i. Therefore, I 1 =I-i Let the resistance of ammeter be R m Applying the Kirchoff’s law to the blue loop below we have V m -V R =0. Therefore V m =V R A I I1I1 i I i I 1 =I-i A VmVm VRVR

7. Current Measurement From the two Kirchoff’s laws I 1 =I-i ; V m =V R Using Ohm’s Law: V m = R m i; V R =R (I-i) Therefore we have the equation R m i = V m =V R = R (I-i) R m i = R (I-i), I i I 1 =I-i A VmVm VRVR

8. Current Measurement Current can be measured using a moving coil ammeter or a digital multi- meter, which is put in series with the circuit. The range of a meter may be extended by inserting a shunt resistance in parallel. If i is the full scale deflection (f.s.d.) current of the meter, R m is the meter resistance and I is the desired new f.s.d. current, then:

9. FP3, questions 5 & 6a (use formula and picture on handout)

10. Voltage Measurement A voltmeter is really an ammeter with a series resistance (multiplier resistance). = V= I(R m +R) R RmRm

11. Voltage Measurement Voltmeter is used in parallel and draws only a small current from the circuit. How to measure voltage across a bulb? No need to cut the circuit, just attach the voltmeter in parallel to the bulb. Has a known FSD (Full Scale Deflection)

12. Voltage Measurement What can we do if voltage is too high: the full scale deflection of the voltmeter is not enough to measure voltage? Let us divide the voltage The range of the meter may be varied using the series resistance.

13. Voltage Measurement What should we do if the voltage is too high? Example: if we want to measure the voltage V 2 across two bulbs, then the voltage will be too high to measure. Indeed, the voltage across the two bulbs is about twice as much as the voltage across one bulb, but even with one bulb only the voltage was nearly the full scale deflection. To divide too high a voltage we can add the resistance in series to our voltmeter. V2V2

14. Voltage Measurement How to estimate the required resistance to extend the range of the voltmeter from v to V ? Applying the Kirchoff’s law to the green loop: V 2 =V =v+V R =v+IR with current flowing through the resistor Since the voltmeter and the resistor are in series, their resistances should add R total = R+R m. Applying Ohm’s Law: I=V/(R+R m ) Substituting the current I to, we arrive to the following equation V=v+RV/(R+R m ) Which we now simplify V 2 =V R I RmRm VRVR Ohm’s Law: V R =IR

15. Voltage Measurement V=v+RV/(R+R m ) Multiplying by R+R m V (R+R m ) = v (R+R m )+RV V R+ V R m = v R+v R m +RV (V -v )R m = v R Dividing by v R=R m (V-v)/v V 2 =V R I RmRm

16. Voltage Measurement A voltmeter is really an ammeter with a series resistance (multiplier resistance). The voltmeter is used in parallel and draws only a small current from the circuit. The range of the meter may be varied using the series resistance. If v is the f.s.d. voltage across the meter, V is the desired new f.s.d. and R m is the meter resistance, then:

17. FP3, questions 4 & 6b (use formula and picture on handout) You will need to think a little for 6b.

18. Potentiometer A potentiometer is a variable potential divider Moveable contact Can be depicted as follows

19. Potentiometer A potentiometer is a variable potential divider What is the resistance between points A and M? Resistance between points A and B can be estimated as: R tot =  L tot /A (1) Resistance between points A and M can be estimated as: : R 1 =  L 1 /A (2) Dividing equation (1) by equation (2) we derive: R tot /R 1 =L tot /L 1  Thus, R 1 =R tot L 1 /L tot  Or R 1 /R tot =L 1 /L tot  - just a simple ratio idea Potentiometer R1R1 R tot L tot A B L1L1 M

20. Potentiometer A potentiometer is a variable potential divider V1V1 V in R tot R1R1 in V out Potential divider Potentiometer R1R1 R tot R 1 =R tot L 1 /L tot ; R 1 /R tot =L 1 /L tot I=V in /R tot ; V 1 =IR 1 =V in R 1 /R tot R tot = R 1 + R 2 V 1 =V in L 1 /L tot

21. Potentiometer A potentiometer is a variable potential divider. In order to change output voltage V 1 we just need to shift a movable contact V1V1 V in R tot R1R1

22. A potentiometer is a variable potential divider which may be used to measure an e.m.f. To measure e.m.f., it is necessary to draw no current though the unknown battery so that there is no voltage loss in the internal resistance. This can be done using a known driver cell, a potentiometer and a galvanometer. Potentiometer ? V 1.5V

23. Potentiometer A potentiometer is a variable potential divider which may be used to measure an e.m.f. To measure e.m.f., it is necessary to draw no current so that there is no voltage loss in the internal resistance. This can be done using a known driver cell, a potentiometer and a galvanometer. The potentiometer is varied until the galvanometer reads zero (null detection) – the e.m.f. of the unknown can then be determined as a fraction of the known e.m.f. 1.5V L1L1 L tot e.m.f.= V 1 = V in L 1 /L tot V1V1 V in e.m.f.

24. Potentiometer A potentiometer is a variable potential divider which may be used to measure an e.m.f. To measure e.m.f., it is necessary to draw no current so that there is no voltage loss in the internal resistance. This can be done using a known driver cell, a potentiometer and a galvanometer. The potentiometer is varied until the galvanometer reads zero (null detection) – the e.m.f. of the unknown can then be determined as a fraction of the known e.m.f.

25. Wheatstone Bridge Potentiometer rearranged to compare two resistances. Wheatstone bridge is a null detection circuit. 1.5V I 0 =0 V 0 =0 Unknown R 3 Known R 4 R1R1 R2R2

26. Wheatstone Bridge The galvanometer reads zero when V 2 =V 4 and V 3 =V 1 This can be seen from the second Kirchoff’s law for red and blue loops and the condition that p.d. between points A and B is zero (the galvanometer reads zero): V 2 +V AB =V 4 V 1 +V AB =V 3 ; V AB =0, thus: V 2 =V 4 and V 3 =V 1 From the first Kirchoff’s law for node A: I 2 =I 1 +I AB with the current I AB between points A and B, but it is zero according to the galvanometer reading. Thus, I 2 =I 1 =I L where I L is the current flowing through the left branch of the bridge. A B I2I2 I1I1

27. Wheatstone Bridge From the first Kirchoff’s law for node B: I 4 +I AB =I 3 with I AB =0, thus, I 4 =I 3 =I R Where I R is the current flowing through the right branch of the bridge A B ILIL ILIL I3I3 I4I4

28. Wheatstone Bridge From the first Kirchoff’s law for node B: I 4 +I AB =I 3 with I AB =0, thus, I 4 =I 3 =I R Where I R is the current flowing through the right branch of the bridge  From Ohm’s Law for all for resistors:  V 1 =I 1 R 1 =I L R 1, V 2 =I 2 R 2 =I L R 2,  V 3 =I 3 R 3 =I R R 3, V 4 =I 4 R 4 =I R R 4  From equations V 2 =V 4 and V 3 =V 1 we derive:  I R R 3 =V 3 =V 1 =I L R 1 I R R 4 =V 4 =V 2 =I L R 1  I R R 3 =I L R 1, thus, I R =I L R 1 /R 3 Substituting I R =I L R 1 /R 3 into I R R 4 =I L R 1 we derive I L R 1 /R 3 x R 4 =I L R 1 then dividing by R 4 Then dividing by R 4 we obtain: A B ILIL ILIL IRIR IRIR

29. Wheatstone Bridge A B ILIL ILIL I3I3 I4I4 Therefore R 1 /R 2 =R 3 /R 4. R 3 is usually unknown resistance, other three are known. In order to balance the bridge, the ratio R 1 /R 2 is varied until galvanometer reads zero. Then R 3 =R 4 (R 1 /R 2 ). ? known

30. Wheatstone Bridge Potentiometer rearranged to compare two resistances. Unknown e.m.f. replaced by a divider that produces a fraction of the known p.d. Wheatstone bridge is a null detection circuit. Reads zero when V 2 =V 4, V 1 =V 3. Therefore R 1 /R 2 =R 3 /R 4. R 3 is usually unknown resistance, other three are known. Ratio R 1 /R 2 is varied until the bridge balances, and galvanometer reads zero. Then R 3 =R 1 R 4 /R 2. The Wheatstone bridge measures resistance. R unknown =R 1 R 4 /R 2 The Wheatstone bridge measures resistance. R unknown =R 1 R 4 /R 2 =

31. This is just simple ratios Wade through presentation 5 if you would like to see the theory Do Q7 Try the other questions on FP3

32. Next week Capacitors Bringing it all together Revision and exam style practice questions Friday – test. (NOT part of your final mark!)

33. If you want to do well… Study! Textbook is very good Ppts are on Learnzone