Download Direct Current (DC) Circuits 1

Direct Current (DC) Circuits 1
PHYS 0212 DC Circuits 1
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Direct Current (DC) Circuits 1
This experiment has four parts:
1. Construct an Ammeter to measure electrical current.
Using a
Galvanometer
{
2. Construct an Ohmmeter to measure electrical resistance.
3. Construct a Voltmeter to measure voltage.
4. Measure the internal resistance of a galvanometer.
Electrical charge is a fundamental property of matter, like mass, and may be
either positive or negative.
The unit of electrical charge is the Coulomb (C).
1.0 C  1.602 1019 electrons
Electrical Current (I) – The amount of charge flowing past a given point per unit
time.
Coulomb
Ampere (amp or A) 
second
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Voltage (V) – The energy per unit of charge.
Joule
Volt (V) 
Coulomb
The Voltage and the Current are related by Ohm’s Law:
V  IR
Volt
R = Resistance, units are:
 Ohm ()
Amp
A circuit is a closed path for electrical current.
+
-
1. Electrons gain potential energy in the battery.
2. They leave the positive end of the battery and
travel to the filament of the light bulb.
3. They lose potential energy in the filament by
converting it to light and heat.
4. The electrons return to the negative end of the
battery and gain more potential energy.
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A circuit is often drawn using a schematic – Lines represent wires and symbols
represent circuit elements.
If the voltage of the batter is V = 1.5 V
and the resistance is R = 100 , what is
the current in the circuit?
V  IR
I
V
I
R
R
1.5 V
I
 0.015 A
100 
+
I
-
I  15 mA
V
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Resistors in a circuit may be connected two different ways: series and parallel.
Series –The current through each device is the same but the voltage change across
each device may be different.
R1
R2
V  IR1  IR2  I  R1  R2 
Rs is a single resistor that can
effectively replace R1 and R2.
IRs  I  R1  R2 
I
IRs  I  R1  R2 
-
+
V
Note that adding resistors
in series increases the
effective resistance.
Rs  R1  R2
In general, if there are N resistors in series,
then the effective resistance for all of them
will be:
Rs  R1  R2  R3 
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 RN
5
Parallel –The voltage across each device is the same but the current through each
device may be different.
R2
V  IRp
I2
Rp is a single resistor that can effectively replace R1
and R2.
R1
V
V V
 
R p R1 R2
I  I1  I 2
I1
I
I
+
V
I
Rp
V
Note that adding resistors
in parallel decreases the
effective resistance.
1
1 1
 
Rp R1 R2
In general, if there are N resistors in parallel, then
the effective resistance for all of them will be:
1
1 1
1
   
Rp R1 R2 R3
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
RN
6
Example: What is the effective resistance of this circuit?
Start with the parallel combination:
1
1
1
1
1
1





Rp R2 R3 200 300 120
R2=200
R1=100 Rs=220 Rp=120
R3=300
Rp  120
Now calculate the series
combination:
Rs  R1  Rp
Rs  100  120
+
Rs  220
V
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What is the effective resistance between points x and y?
300 
2000 
600 
x
y
500 
200 
a) 80 
b) 3600 
c) 764 
d) 500 
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Galvanometer
A device that measures electrical current in a circuit.
A galvanometer must be placed in series in a circuit so
that the current passes through the meter. For this
reason the internal resistance of the galvanometer Rm
should be as small as possible, ideally zero ohms.
G
Rm
R
The arrow
means the
resistor R is
variable.
+
G
Rm
This is the basic circuit you will use for the first
three parts of the lab. Ohm’s law for this
circuit is:
V  I  R  Rm 
-
V
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The power through the resistance
substitution box must not exceed
2 Watts.
P  IV
(2 W)  I  5 V 
I  0.40 A  400 mA
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V  IR
V
5V
R 
I 0.4 A
R  13 
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A word of advice.
Adjust the resistance on the decade box to get an even value of the current.
2660
2497 
It is difficult to estimate the value of the current between the tick marks. However,
if you adjust the resistance so the needle falls right on a tick mark each time then
your results will be much more accurate.
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Construction of an Ammeter
An ammeter measures electrical current I.
V  I  R  Rm 
R
Solve Ohm’s law for R.
G
R
V
 Rm
I
Rm
+
-
Make the substitution:
x 1 I
R  Vx  Rm
V
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R  Vx  Rm
Make a plot of R versus x.
slope  V
intercept   Rm
The voltage of the power supply.
The internal resistance of the
galvanometer.
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Construction of an Ohmmeter
An ohmmeter measures electrical resistance R.
Same equation as before: R  Vx  Rm
The values of V and Rm come form part
one.
R
G
Rm
+
V
-
In this part of the lab you will put
known values of resistance into the
circuit and use this equation to
calculate R.
You will use this circuit to measure
individual resistors as well as series
and parallel combinations of resistors.
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Construction of an Voltmeter
An voltmeter measures voltage V.
The voltage may be determined with this
equation:
V  I  R  Rm 
R
G
Rm
+
-
V
Say I  2 mA
and R m  10
The value of Rm comes form part one and the
resistance R is preset.
In this part of the lab you will put known
sources of voltage into the circuit and use this
equation to calculate V.
The preset value of R may be determined using
Ohm’s law:
If Vmax  30 V
and I max  5 mA
then V   2 10-3 A   5990  10 
V  12.0 V
then R 
Vmax
30 V
 Rm 
 10  5990
-3
I max
5 10 A
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The Internal Resistance of the Galvanometer
You will determine the internal resistance of the galvanometer Rm from the
intercept of R versus x (1/I), but how accurate is it? This procedure will determine
Rm more directly.
Step 1 – Adjust the external resistor
R1 until the current through the
galvanometer is 4 mA.
R1
V
IA 
R1  Rm
G
Rm
+
V
R2
-
Step 2 – Add a second external
resistor R2 in parallel with the
galvanometer and adjust it until the
current through the galvanometer
is 2 mA.
V
IB 
 Rm R2 
R1  

R

R
2 
 m
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Rm R2
1

1
1
Rm  R2
Rm  R2
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Assume that R2  Rm  10, R1  1000  and that V  5 V.
IA 
V
5V
5V


 4.00 mA
R1  Rm 1240  10 1250
IB  I A
V
V
5V
5V
IB 



 4.02 mA
1
 Rm Rm  R1  2 Rm 1240  5 1245
R1  

R

R
m 
 m
This tells us that when we add R2 to the circuit the total current stays
approximately the same (only 0.4% difference).
So basically, we can turn this around and say that when we adjust R2 such that
the current through the galvanometer is cut in half then:
R2  Rm
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