AIM: to determine the behavior of an inductive resistivity of an A.C circuit.
APPARATUS: low voltage AC supply(10-12 volts). Fixed inductance (0.1HZ) with negligible distance variable non inductive resistance (standard plug- in box with 5-ohm steps). A.C ammeter with range0.1A, A.C voltmeter and key switch.
Theory ;
For the circuit shown in the diagram above,
I=V/Z =V/
X is called the inductive reactance of the coil and it is given as (2∏L)
Where F= frequency of mains supply=60 circles per seconds. L= inductance of the coil.
It therefore follows that
Z^2 =R^2 + X^2
Thus creating a graph of Z^2 against R^2 will yield a straight line as the figure below
The intercept OA =x^2 =(2∏FL)^2
Hence, from the inductance of the coil maybe calculated
L=(OA)1/2÷2∏F
PROCEDURE
Plug in the power source to the mains and connect the low voltage output as shown in figure 23a above with the key switch open. When you have finished, connect out the circuit (ensure to ask the instructor to check the circuit). Then check the readings of the voltmeter which is constant and record its value, when the key is closed also do the same. Then vary R in the steps of 5 ohms and record the current reading of the ammeter.
Take at least six readings and tabulate your result as follow
R
R^2
I
Z=V/I
Z^2
below are the values if you can't see the table very well.
50
2500
2.88
5.21
27.14
100
10000
2.76
5.43
29.48
150
22500
2.64
5.68
32.26
200
250
40000
62500
2.52
2.46
5.95
6.09
37.41
37.08
300
90000
2.34
6.41
41.08
CALCULATIONS
PRECAUTIONS
1. I ensure that the switch was open before each step is completed
2. I avoided error due to parallax by placing my eyes vertically downward
3.I ensure that the circuit was connected properly before taking my readings.
GRAPH
Note; dont try to form too smart, the value you see on the graph is the highest we saw so far...unless you know what you are doing.
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