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Explanation & Answer
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CAPACITIVE REACTANCE
Objectives
At the end of the experiment one should be able to measure a capacitor’s reactance at a
given frequency. Also they should make a comparison between capacitors’ reactance in parallel or
series connections.
Introduction
There is flow of current in a phase when some there is some applied voltage and with a
connection of a resistor across a sine wave generator. When a capacitor is connected across the
generator instead of a resistor, the current fails to be in phase with the voltage. The flow of current,
the capacitance and the rate at which voltage changes in the capacitor are all directly proportional.
If there is a maximum change of voltage, the current flow is largest. More current flows once
capacitance or frequency are increased, creating the thought of a capacitor as a high-frequency
short.
Similar to resistance in ac current flow is reactance which is measured in ohms. Capacitive
reactance is expressed as XC, XC = (1/2πf C), where f is the frequency in hertz and C is the
capacitance in farads. In ac circuits, the Ohm’s law can be generalized. The voltage across a
capacitor can be determined using both capacitive reactance and current in the capacitor. The law
can be expressed as VC = IXC
Procedure
Materials
Capacitors: one of each 0.1 µF, 0.047 µF, 1.0 µF
Resistors: one of each 1.0 kΩ, 4.7 kΩ, 10 kΩ.
Two 100 µF capacitors, two LEDs, and one 100kΩ resistor
Procedure
I obtained two capacitors and drew a table labelled it as shown. Since there were a
capacitance bridge, I measured and recorded their capacitance. Similarly one can record the values
of the given capacitors. Then I measured and recorded resistor R1’s reading. I set the circuit as
required and connected the generator for 1 kHz sine wave with an output of 1.0V rms. With the
DMM still in connected, I measured the rms voltage and checked both the frequency and voltage
using the oscilloscope. And 1.0 V = 2.828 pp. I connected the circuit in series to ensure that resistor
was identical to the current by the capacitor. Through application of the Ohm’s law to the resistor,
one can find the current. Using DMM I measured the voltage VR across the resistor and recorded
it in C1. By diving the voltage by the R1’s resistance, I then computed current and recorded in table
C (2).
The other step was to measure and record the rms voltage across the capacitor VC. I then
used the Ohm’s law with this voltage to calculate the capacitive reactance and entered it in table
C (2). XC = VC/I. Using this capacitive reactance I computed the capacitance according to the
equation: C = 1/ 2π f XC and recorded them table C (2). The value was supposed to be equal with
the measured value of the capacitor and that marked on it. Using the C2 capacitor I repeated the
previous steps and recorded the results at C2.
The next step was to connect capacitor C1 in series with C2. Both the capacitive reactance
and capacitance in the capacitor are equivalent and I determined them by measuring the capacitors
across the series as a single capacitor. I then recorded the results in Series Capacitors. I used the
following steps: checked if the generator was set to 1.0 V rms and determined the circuit’s current
by dividing the measured voltage by resistance and recorded them in table C (3). Second I
measured the voltage across both capacitors and recorded them. Then using Ohm’s law I
determined the capacitive reactance of the two capacitor using the measured current and voltage
in steps 1 and 2. Lastly, I computed the total capacitance by the equation CT = 1/ 2π f XC
I repeated the list of above steps with the capacitors connected in parallel assuming that
they are equivalent to one for measurements. I recorded the results at Parallel Capacitors.
Data analysis
Table C (2)
Component
C1
C2
R1
Listed value
0.1 µ F
0.047 µ F
1.0 k Ω
Measured value
0.186V
0.058 µV
0.972V
Capacitor C1
1.4 V pp
1.4 mA
2.25 V pp
1.6 k Ω
6.1064 µ
Capacitor C2
1.9 V pp
1.35 mA
3.25 V pp
1.67 k Ω
6.0 µ
Table...