1. Measuring Instruments: Voltmeter You can measure a voltage by placing a galvanometer in parallel with the circuit component across which you wish to measure the potential difference. V=3 V R=10  r=0.5  G RGRG a b V ab =?

2. Example: an galvanometer of resistance 60  is used to measure the voltage drop across a 10 k  resistor in series with a 6 V battery and a 5 k  resistor (neglect the internal resistance of the battery). What is the percent error caused by the nonzero resistance of the galvanometer? First calculate the actual voltage drop. V=6 V R 1 =10 k  R 2 =5 k  a b

3. The measurement is made with the galvanometer. V=6 V R 1 =10 k  R 2 =5 k  G R G =60  a b 60  and 10 k  resistors in parallel are equivalent to an 59.6  resistor. The total equivalent resistance is 5059.6 , so 1.19x10 -3 A of current flows from the battery. I=1.19 mA The voltage drop from a to b is then measured to be 6-(1.19x10 -3 )(5000)=0.07 V. The percent error is. Your opinions? Would you pay for this voltmeter?

4. To reduce the percent error, the device being used as a voltmeter must have a very large resistance, so a voltmeter can be made from galvanometer in series with a large resistance. V G RGRG R Ser  Everything inside the blue box is the voltmeter. a b V ab a b Homework hints: “the galvanometer reads 1A full scale” would mean a current of I G =1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected R Ser must result in I G =1A when V ab =10V. Homework hints: “the galvanometer reads 1A full scale” would mean a current of I G =1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected R Ser must result in I G =1A when V ab =10V.

5. Example: a voltmeter of resistance 100 k  is used to measure the voltage drop across a 10 k  resistor in series with a 6 V battery and a 5 k  resistor (neglect the internal resistance of the battery). What is the percent error caused by the nonzero resistance of the voltmeter? We already calculated the actual voltage drop (3 slides back). V=6 V R 1 =10 k  R 2 =5 k  a b

6. The measurement is now made with the voltmeter. V=6 V R 1 =10 k  R 2 =5 k  V R V =100 k  a b 100 k  and 10 k  resistors in parallel are equivalent to an 9090  resistor. The total equivalent resistance is 14090 , so 4.26x10 -4 A of current flows from the battery. I=.426 mA The voltage drop from a to b is then measured to be 6-(4.26x10 -4 )(5000)=3.9 V. The percent error is. Not great, but much better. Larger R ser is needed for high accuracy.

7. An ohmmeter measures resistance. An ohmmeter is made from a galvanometer, a series resistance, and a battery. G RGRG R Ser R=? The ohmmeter is connected in parallel with the unknown resistance with external power off. The ohmmeter battery causes current to flow, and Ohm’s law is used to determine the unknown resistance.  Measuring Instruments: Ohmmeter Everything inside the blue box is the ohmmeter.

8. To measure a really small resistance, an ohmmeter won’t work. Solution: four-point probe. V A reference: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movcoil.html#c4http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movcoil.html#c4 Measure current and voltage separately, apply Ohm’s law.