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A. Ammeter: • measures current (A) • connected in series. (current must go through instrument)
Typology: Study notes
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Today’s agenda:
Measuring Instruments: ammeter, voltmeter, ohmmeter. You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments.
RC Circuits. You must be able to calculate currents and voltages in circuits containing both a resistor and a capacitor. You must be able to calculate the time constant of an RC circuit, or use the time constant in other calculations.
Ammeter:
a b
Voltmeter:
Ohmmeter:
Measuring current in a simple circuit:
R +r
To minimize error, ammeter resistance r must be very small. (ideal ammeter would have zero resistance)
r
Are we measuring the correct current? (the current in the circuit without ammeter)
Example: an ammeter of resistance 10 mΩ is used to measure the current through a 10 Ω resistor in series with a 3 V battery that has an internal resistance of 0.5 Ω. What is the relative (percent) error caused by the ammeter?
r=0.5 Ω
Actual current without ammeter:
V I = R +r
3 I = A 10 +0.
I = 0.2857 A = 285.7 mA
You might see the symbol ε used instead of V.
Designing an ammeter
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c
Galvanometer:
Designing an ammeter
Needle deflection is proportional to current. Each galvanometer has a certain maximum current corresponding to full needle deflection.
What if you need to measure a larger current?
Shunt also reduces resistance of the ammeter:
A G SHUNT
G SHUNT A G SHUNT
A galvanometer-based ammeter uses a galvanometer and a shunt, connected in parallel:
A G S
Example: what shunt resistance is required for an ammeter to have a resistance of 10 mΩ, if the galvanometer resistance is 60 Ω?
S A G
( ) ( ) S =^ G^ A =^ =^ Ω G A
R - R 60 -.01 (^) (actually 0.010002 Ω)
To achieve such a small resistance, the shunt is probably a large-diameter wire or solid piece of metal.
Measuring voltage (potential difference) Vab in a simple circuit:
Are we measuring the correct voltage? (the voltage in the circuit without voltmeter)
ε=3 V
r=0.5 Ω
a (^) b
To minimize error, voltmeter resistance r must be very large. (ideal voltmeter would have infinite resistance)
Measuring voltage (potential difference) Vab in a simple circuit:
Are we measuring the correct voltage? (the voltage in the circuit without voltmeter)
ε=3 V
r=0.5 Ω
a (^) b
The measurement is made with the galvanometer.
R 1 =10 kΩ
R 2 =5 kΩ
a (^) b
60 Ω and 10 kΩ resistors in parallel are equivalent to 59.6 Ω resistor.
Total equivalent resistance: 5059.6 Ω
Total current: I=1.186x10-3^ A
I=1.19 mA
Vab = 6V – IR 2 = 0.07 V.
The relative error is:
×
% Error = 100 = 98% 4 Would you pay for this voltmeter?Would you pay for this voltmeter? We need a better instrument!
Example: a voltmeter of resistance 100 kΩ is used to measure the voltage drop across a 10 kΩ resistor in series with an ideal 6 V battery and a 5 kΩ resistor. What is the percent error caused by the nonzero resistance of the voltmeter?
We already calculated the actual voltage drop (2 slides back).
R 1 =10 kΩ
R 2 =5 kΩ
a (^) b
RSer^ RG ⇒
Everything inside the blue box is the voltmeter.
a (^) b
Vab
a (^) b
Vab
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 Vab=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 Vab=10V.
RSer^ RG
Ω
Everything inside the blue box is the ohmmeter.