Download Electricity

Electricity
RLC circuits
Discharging a capacitor
Charging a capacitor
RC circuits and square shaped voltage
Phase shift between voltage and current through capacitor
Phase shift between voltage and current through resistor and capacitor in series
RC circuits as "frequency filters"
RL circuit and square shaped voltage
Phase shift between voltage and current through inductor
Oscillations of RLC circuit
Forced oscillations of RLC circuit
Discharging capacitor
DC voltage V0 is connected to an ideal capacitor and then voltage source is removed. If resistance between
both plates of the capacitor would be infinite, the voltage on a capacitor VC would keep constant value equal
to V0. If a capacitor C is connected to a resistor R, voltage on the capacitor decreases in time following the
equation:
.
Product RC is called time constant .
Performing the experiment
Apparatus: resistor 20 to 200 k, capacitor 1 to 100 F, two switches.
Connect the capacitor, resistor and two switches to e-ProLab interface as shown.
Start the measurement, press S1 for about a second then release S1 and immediately press S2 holding it
down until the end of sampling.
Example values: R=47 k, C=47 F (electrolytic)
Calibrated e-ProLab file
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Typical results:
Example curve obtained at R=47 k, C=47 F
Example e-ProLab file obtained at R=47 k, C=47 F
Analyses
Export data to an analyses software (say a spreadsheet programme) and determine time constant . Taking
into account, that tolerance of resistors is smaller then for capacitors, calculate capacitance and compare it
with its declared value.
Example Excel file.
Variations
1. Keep the same C, vary R (R in parallel and in series)
2. Keep the same R, vary C (C in parallel and in series)
3. Keep the same C, use light dependent resistor at different illuminations
4. Keep the same C, use NTC thermistor kept at different temperatures
Charging capacitor
Say that first there is no electric charge on the capacitor. Then DC voltage V O is connected to an ideal
capacitor in series with the resistor R. If capacitor C is connected to a resistor R. Voltage on the capacitor
increases in time following the equation:
.
Performing the experiment
Apparatus: resistor 20 to 200 k, capacitor 1 to 100 F.
Analogue output Vout(1) of the interface can be used as voltage source V O. Connect capacitor, resistor and
two switches to e-ProLab interface as shown:
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Hold the switch S1 down. Start the sampling, release S1 and immediately press switch S2 holding it down
until the sampling stops.
Example values: R=47 k, C=47 F (electrolytic)
Calibrated e-ProLab file
Typical results:
Example curve obtained at R=47 k, C=47 F
Example e-ProLab file obtained at R=47 k, C=47 F
Variations
1. Keep the same C, vary R (R in parallel and in series)
2. Keep the same R, vary C (C in parallel and in series)
3. Keep the same C, use light dependent resistor at different illumination
4. Keep the same C, use NTC thermistor kept at different temperatures
RC circuits and square shaped voltage
We can charge and discharge a capacitor by generating step-up output voltage signal to charge the
capacitor and step-down voltage to discharge the capacitor. Consequently, a square shape of the output
signal does both.
Performing the experiment
Apparatus: resistor 2 to 200 k or light dependent resistor or thermistor, capacitor 1 to 100 F.
Analogue output Vout2 of the interface can be used to generate square output function alternating between
0V and chosen positive voltage or between negative and negative voltage.
Connect a capacitor and resistor to e-ProLab interface as shown:
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Start sampling and compare curves of voltage on the capacitor and current with the output voltage curve.
Example values: R=Light dependent resistor, C=2.2 F.
Calibrated e-ProLab file, Vout2 is square shaped voltage alternating between +5V and -5V, frequency
is 3,3 Hz, top chart is Vout2 and voltage on the capacitor, bottom chart is Vout2 and current.
Typical results:
Example curve obtained at LDR declined by a hand, C=2.2 F, Vout2 is square shaped voltage
alternating between +5V and -5V, frequency is 3,3 Hz, top chart is Vout2 and voltage on the capacitor,
bottom chart is Vout2 and current.
Example e-ProLab file obtained at LDR, C=2.2 F, Vout2 is square shaped voltage alternating
between +5V and -5V, frequency is 3,3 Hz, top chart is Vout2 and voltage on the capacitor, bottom chart is
Vout2 and current.
Variations
1. Vary illumination of the LDR
2. Keep the same C, use NTC thermistor kept at different temperatures
Phase shift between voltage and current through capacitor
A capacitor and a resistor in series are connected to a generator of sine voltage Vout2 as shown in scheme
below. Voltage vs. time on the resistor Vin6 is sampled. Voltage across the capacitor VC=Vout2-Vin6 and
current through both of them is obtained according to the Ohms law I=Vin6/R.
Investigating VC and I vs. time, the following relations can be investigated:
 Phase shift between current and voltage
 Relation between amplitude of voltage at the capacitor and amplitude of current is called absolute
impedance:
Performing the experiment
Apparatus: resistor 100  to 10 k, capacitor 0.4 to 10 F.
Connect the capacitor, resistor to the e-ProLab interface. Analogue output Vout2 is used as sine voltage
generator, analogue input Vin6 is sampling voltage across the resistor. Software needs to plot current curve
I(t) obtained from Vin6 and voltage across the capacitor VC(t) = Vout2-Vin6.
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Start sampling and observe both curves on the same chart.
Example values: R= 1k, C=2.2 F, f=50 Hz
Calibrated e-ProLab file for R= 1k, f=50 Hz, amplitude of Vout is 10 V, displaying VC(t) and I(t) on
one chart, VC vs. I on other chart.
Typical results:
Example curves obtained at R= 1k, C=2.2 F; f=50 Hz and f=250 Hz
Example e-ProLab file obtained at R= 1k, C=2.2 F, f=50 Hz
Analyses
1. Determine phase shift of voltage curve VC with regard to current curve I.
2. Keep the same C and R, vary frequency f, investigate how Xc depends on f
3. Keep the same R and f, vary C, investigate how Xc depends on C
Phase shift between voltage and current through resistor and capacitor in series
A capacitor and a resistor in series are connected to a generator of sine voltage Vout2 as shown in scheme
below. Voltage vs. time on the resistor Vin6 is sampled. Current through both of them is obtained according
to the Ohms law I=Vin6/R.
Investigating Vout2 and I vs. time, the following relations can be investigated:
 Phase shift between current and voltage
 Relation between amplitude of output voltage and amplitude of current is called absolute
impedance:
Performing the experiment
Apparatus: resistor 100  to 10 k, capacitor 0.4 to 10 F.
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Connect the capacitor, resistor to the e-ProLab interface. Analogue output Vout2 is used as sine voltage
generator, analogue input Vin6 is sampling voltage across resistor. Software needs to plot current curve I(t)
obtained from Vin6 and generator's voltage Vout2.
Start sampling and observe both curves at the same chart.
Example values: R= 1k, C=2.2 F, f=50 Hz
Calibrated e-ProLab file for R= 1k
Calibrated e-ProLab file for R= 10k
Typical results:
Example curves obtained at C=2.2 F, f=50 Hz; R= 1kand R= 10k
Example e-ProLab file obtained at C=2.2 F, f=50 Hz; R= 1k
Analyses
1. Keep the same C and R, vary frequency f, investigate how Z depends on f
2. Export data to spreadsheet software and plot power curves for the generator, power at the resistor and at
the capacitor; calculate average power supplied by the generator and average power at the resistor and at
the capacitor during one period of the output signal (not the first period!).
Example Excel file.
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RC circuits as "frequency filters"
The circuit consists of R and C in series connected to AC generator. Say that generator's amplitude is fixed
while its frequency is varied. The voltage amplitude on the resistor and on the capacitor depends on
frequency. Two corresponding circuits can be studied:
 voltage amplitude on the resistor is sampled and compared to generator's voltage (Filter 1 on the
figure below) and
 voltage amplitude on the capacitor is sampled and compared to generator's voltage (Filter 2 on the
figure below).
Both circuits can be used as frequency filters. Filter circuits are classified in four groups: low-pass,
high-pass, band-pass and notch filters.
Performing the experiment
Apparatus: resistor 100  to 10 k, capacitor 0.4 to 10 F.
Connect the capacitor, resistor to the e-ProLab interface. Analogue output Vout2 is used as AC voltage
generator, analogue input Vin6 is sampling voltage across resistor. Software needs to plot current curves
Vout2 and Vin6.
Filter 1,
Filter 2
Start with Filter 1. Form sine Vout2 voltage. Fix the amplitude of Vout and compare the amplitude of Vin6
and Vout2 at different frequencies. Find the two example frequencies; f1 where amplitude of Vin6 is not
significantly smaller than Vout2 (say that amplitude 0.85 < Vin6/Vout2 < 0.95) and f2 where amplitude of
Vin6 is significantly smaller than Vout2 (say that amplitude 0.05 < Vin6/Vout2 < 0.15).
Finally form Vout2 signal to be a sum (superposition) of sine signal having frequency f1 and signal having
frequency f2. Note that both amplitudes must not exceed the range of Vout2 (-10 V to +10V for the interface
CMC-S3). Compare the sampled curve with the generator's curve and consider what would be adequate
term for Filter 1; low-pas of high-pass?
Finally, replace positions of R and C in the circuit to make the circuit Filter 2. Let Vout2 generate the same
superposed form (two sine components with the same frequencies f1 and f2 as determined for Filter 1 circuit.
Compare the sampled curve with the generator's curve and consider what would be adequate term for
Filter 1; low-pas of high-pass?
Example values: R= 1k, C=2.2 F, f1=800 Hz, f2=20 Hz
Calibrated e-ProLab file for superposition of f1=800 Hz (amplitude 3V) and f2=20 Hz (amplitude 7V)
Typical results:
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Example time charts and Fourier transform obtained with Filter 1, Vout2 top chart, Vin6 bottom chart,
C=2.2 F, R= 1kf1=800 Hz (amplitude 3V) and f2=20 Hz (amplitude 7V)
Example time charts and Fourier transform obtained with Filter 2, Vout2 top chart, Vin6 bottom chart,
C=2.2 F, R= 1kf1=800 Hz (amplitude 3V) and f2=20 Hz (amplitude 7V)
Example e-ProLab file obtained with Filter 1, Vout2 top chart, Vin6 bottom chart, C=2.2 F, R=
1kf1=800 Hz (amplitude 3V) and f2=20 Hz (amplitude 7V) , Fourier transform window is pre-set and may
be open.
Example e-ProLab file obtained with Filter 2, Vout2 top chart, Vin6 bottom chart, C=2.2 F, R=
1kf1=800 Hz (amplitude 3V) and f2=20 Hz (amplitude 7V) , Fourier transform window is pre-set and may
be open.
Analyses
1. If having some theoretical background, calculate cut-off frequency for given RC filter.
RL circuit and square shaped voltage
Square shaped voltage applied on an inductor in series with a resistor causes self induction effect on the
inductor.
Performing the experiment
Apparatus: resistor 2 to 50 k, inductor with iron core to have inductance > 1 H (could be primary coil of a
transformer).
Analogue output Vout2 of the interface can be used to generate square output function alternating between
positive and negative voltage. Connect an inductor and resistor to e-ProLab interface as shown:
Start sampling and observe both curves at the same chart.
Example values: R= 10 k, L =primary coil of a transformer with an ohmic resistance of 60 220 V to 2 x
18 V, 30 VA) Vout is square shaped alternating between +5V and -5V, frequency is 3,3 Hz,
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Calibrated e-ProLab file, Vout is square shaped alternating between +5V and -5V, frequency is 3,3
Hz, top chart is Vout2 and voltage on the inductor, bottom chart is Vout and current.
Example results for R=10 kL=primary coil of the transformer; top chart is Vout2 (brown) and
voltage on the inductor, bottom chart is Vout and current.
Example e-ProLab file obtained with R=10 kL=primary coil of the transformer; top chart is Vout2
(brown) and voltage on the inductor, bottom chart is Vout2 and current.
Phase shift between voltage and current through inductor
An inductor and a resistor in series are connected to a generator of sine voltage Vout2 as shown in scheme
below. Voltage vs. time on the resistor Vin6 is sampled. Voltage across the inductor VL=Vout2-Vin6 and
current through both of them is obtained according to the Ohms law I=Vin6/R.
Investigating VL and I vs. time, the following relations can be investigated:
 Phase shift between current and voltage
 Relation between amplitude of voltage at the capacitor and amplitude of current is called absolute
impedance:
Results obtained with an inductor with iron core (say primary coil of a transformer) show significant deviation
from the expected ideal results. The main reason is nonlinearity caused by ferromagnetism (hysteresis loop)
of the iron core.
Inductors without iron core have much smaller inductance L and require therefore higher driving frequencies.
Performing the experiment
Apparatus: resistor 2 k to 50 k, inductor with iron core to have inductance > 1 H (could be primary coil of
a transformer).
Connect the inductor and resistor to the e-ProLab interface. Analogue output Vout2 is used as sine voltage
generator, analogue input Vin6 is sampling voltage across the resistor. Software needs to plot current curve
I(t) obtained from Vin6 and voltage across the inductor VL(t) = Vout2-Vin6.
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Start sampling and observe voltage curve on the capacitor.
Example values: R= 10 k, L =primary coil of a transformer with an ohmic resistance of 60 220 V to 2 x
18 V, 30 VA), Vout2 is sinusoidal with amplitude 10 V and frequency 50 Hz,
Calibrated e-ProLab file, Vout is sinusoidal, amplitude 10 V, frequency 50 Hz, current I and voltage at
the inductor VL are plotted as well as how VL depends on I.
Example results for R=10 kL=primary coil of the transformer; current and voltage at the inductor
are plotted as well as how VL depends on I.
Example e-ProLab file obtained with R=10 kL=primary coil of the transformer; current and voltage
at the inductor are plotted as well as how VL depends on I.
Analyses
1. Determine phase shift of voltage curve VL with regard to current curve I.
2. Keep the same L and R, vary frequency f, investigate how XL depends on f
Oscillations of RLC circuit
Discharging a capacitor trough an inductor causes oscillations of voltage and current. The characteristic
frequency f0 of oscillation depends on the capacitance and inductance as follows:
Since capacitors and especially inductors do not have ideal characteristics, the oscillation in LC circuit is
significantly damped. Ohmic resistance of the inductor is the most influencing reason for damping. If the
inductor has an iron core, the ferromagnetic behaviour of iron (hysteresis loop and eddy currents) causes
additional damping. Moreover, frequency f0 depends on amplitude (nonlinearity of B-H relation).
Voltage vs. time on the capacitor Vin6 is sampled.
Performing the experiment
Apparatus: capacitor 1 F to 10 F, inductor with iron core to have inductance > 1 H (could be primary coil
of a transformer).
Connect the inductor and capacitor to the e-ProLab interface. Analogue output Vout2 is used to initially
charge the capacitor by generating DC voltage. At the moment of starting the sampling, Vout2 decreases
voltage to 0V to enabling discharging of the capacitor. Analogue input Vin6 is sampling voltage across the
capacitor. Software needs to plot voltage curve obtained from Vin6.
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Start sampling and observe both curves at the same chart.
Example values: R= 2.2 F, L =primary coil of a transformer with an ohmic resistance of 60 220 V to 2 x
18 V, 30 VA).
Calibrated e-ProLab file, Vout2 is initially 10 V and changed to 0 V when sampling is started, voltage
at the capacitor is plotted (click start button two times if first time there are no oscillations).
Example results for R= 2.2 F, L =primary coil of a transformer, Vout2 is initially 10 V and changed to
0 V when sampling is started; voltage at the capacitor is plotted.
Example e-ProLab file obtained for R= 2.2 F, L =primary coil of a transformer, Vout2 is initially 10 V
and changed to 0 V when sampling is started; voltage at the capacitor is plotted.
Analyses
1. Determine characteristic frequency f0.
2. Calculate inductance L of the inductor from known C and f0
3. Keep the same inductor, vary C.
Forced oscillations of RLC circuit
If RLC circuit in series is connected to sine generator, the system cannot oscillate with characteristic
frequency f0
but it oscillates with the frequency f that is "forced" by the generator.
The impedance of the circuit is as follows:
If f>>f0, the impedance of inductor dominates and amplitude of voltage on the inductor is much bigger than
at the capacitor. On the contrary, if f<<f0 , impedance of inductor dominates and amplitude of voltage on the
capacitor is much bigger than at the inductor.
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At f=f0, the impedance of the system is minimal; the circuit is in resonance. The amplitudes at the capacitor
and inductor are equal.
Because of nonlinear behaviour of the iron core in the inductor, the resonance frequency depends slightly on
the amplitude of forced oscillations.
Performing the experiment
Apparatus: capacitor 1 F to 10 F, inductor with iron core to have inductance > 1 H (could be primary coil
of a transformer).
Connect the inductor and capacitor to the e-ProLab interface. Analogue output Vout2 is used to generate
sinusoidal voltage to force oscillations. Voltage on the capacitor VC is sampled through Vin6 while voltage on
the inductor VL=Vout2-Vin6. Both curves, VC and VL are compared to forced voltage Vout2.
Start sampling and observe the curves on the same chart.
Example values: R= 2.2 F, L =primary coil of a transformer with an ohmic resistance of 60 220 V to 2 x
18 V, 30 VA).
Calibrated e-ProLab file, the amplitude of Vout2 is 4V, forced frequencies are f=3,33 HZ,
6,0 Hz and 40 Hz, brown curve is Vout2, blue is VC, light blue is VL.
Example results for R=2.2 F, L =primary coil of a transformer, the amplitude of Vout2 is
4V, forced frequencies are f=3,33 HZ, 6,0 Hz and 40 Hz, brown curve is Vout2, blue is VC, light blue is VL.
Example e-ProLab file obtained for R= 2.2 F, L =primary coil of a transformer, the
amplitude of Vout2 is 4V, forced frequencies are f=3,33 HZ, 6,0 Hz and 40 Hz, brown curve is Vout2, blue is
VC, light blue is VL.
Analyses
1. Find characteristic frequency f0.
2. Determine phase shift of VC and VL regarding Vout2 at characteristic frequency f0, at f<f0 and at f> f0.
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