6.2.2: Explorations

Exploration 1: Amplitude, Frequency, and Phase Shift

We characterize the voltage (or current) in AC circuits in terms of the amplitude, frequency (period), and phase. The sinusoidal voltage of this function generator is given by the equation

\[V (t) = V_{0} \sin(\omega t -\phi ) = V_{0} \sin(2\pi ft -\phi )\nonumber\]

where \(V_{0}\) is the amplitude, \(f\) is the frequency (\(\omega  = 2\pi f\) is the angular frequency), and \(\phi\) is the phase angle (voltage is given in volts and time is given in seconds). Restart.

To begin with, keep the resistance of the variable resistor equal to zero. Pick values for the voltage amplitude (between \(0\) and \(20\text{ V}\)), frequency (between \(100\) and \(2000\text{ Hz}\)), and phase angle (between \(-2\pi\) and \(2\pi\)).

1. What does the amplitude on the graph correspond to?
2. If you increase the amplitude, what do you expect to happen? Try it.
3. Measure the time between two peaks (or valleys) on the graph. This is the period (\(T\)). What does \(1/T\) equal?
4. What do you need to change to increase the time between two peaks? Try it.
5. Compare the plots when \(\phi = 0\) and when \(\phi = 0.5\ast\pi\). (You can right-click inside the plot to make a copy.)
6. What happens when \(\phi =\pi\)?
7. Pick a value of \(\phi\) other than \(0\). Measure the time, \(t\) (measured from \(t = 0\)), it takes for the graph to cross the horizontal axis with a positive slope (going up). \(\phi\) should be equal to \(2\pi ft\). So, the phase (or phase shift) tells you how much the graph is shifted from a straight \(\sin 2\pi ft\) curve.
8. Note that when \(\phi = 0.5\ast\pi\), the plot is a cosine curve. Why?

Now, change the variable resistor. The plot shows both the voltage across the \(1000\:\Omega\) resistor \(\color{blue}{\text{(blue)}}\) and the voltage supply \(\color{red}{\text{(red)}}\). Kirchhoff's laws hold for any instant of time in an AC circuit.

9. Use the techniques you learned for DC circuits to calculate the current in the circuit at several different points.
10. Verify that this circuit is simply a voltage divider.
11. What value does the variable resistor need to have for the maximum voltage across the \(1000-\Omega\) resistor to be \(1/3\) of the value of the source?

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian.

Exploration 2: Reactance

Assume an ideal power supply. For a resistor, we call the ratio between peak voltage and current, the resistance, \(R\). For capacitors and inductors, we use the term the reactance, \(X\), for the ratio between peak voltage and current so that \(V = I X\). This Exploration shows that for an active load like a capacitor or inductor, the reactance depends on the frequency as well (voltage is given in volts, current is given in amperes or milliamperes (note graph labels), capacitance is given in farads, inductance is given in henries, and time is given in seconds). Restart.

1. For a capacitive load, vary the frequency and observe what happens to the current. How is this result related to the formula for capacitive reactance? The graph shows the voltage across \(\color{red}{\text{(red)}}\) and current from the power supply \(\text{(black)}\) as a function of time. Note that you may need to wait for transient effects to decay if you change the frequency.
2. Double the capacitance and try it again. What happens? Explain your observations in terms of the capacitive reactance.
3. Repeat a) and (b) with an inductive load. What happens if you double the inductance? Explain your observations in terms of the inductive reactance.

If we take this to the limit of \(f\to 0\) (DC circuits), then a capacitor is essentially an open circuit. At high frequencies, the capacitor is essentially a short circuit (acts like a wire with little or no resistance).

4. Explain this in terms of what a capacitor does (stores charge) and how it works.
5. At low frequencies, is an inductor essentially an open circuit or a short circuit? What about at high frequencies?
6. Explain in terms of what an inductor does (in terms of induced current).

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Exploration 3: Filters

Since the reactance varies with frequency, we can use capacitors (or inductors) to filter out different frequencies. The voltage of the source is \(\color{red}{\text{red}}\), while the oscilloscope voltage is the \(\color{blue}{\text{blue}}\) plot on the graph (voltage is given in volts and time is given in seconds).

1. At very low frequencies does the capacitor have a high or a low reactance?
2. Therefore, at low frequencies, will the current through the capacitor be large or small?

Capacitor filter: Try filter 1.

3. Will this circuit allow high or low frequency signals to reach the oscilloscope? Explain.
4. Try it.
5. Is the amplitude of the voltage measured by the oscilloscope bigger at low or high frequencies?

If it is bigger at higher frequencies, then it "allows" high frequencies through more readily than lower frequencies and it is called a high-pass filter. If it "allows" low frequencies through, it is a low-pass filter. Look at the circuit for filter 2.

6. Do you think this is a high-pass or low-pass filter? Why?
7. Try it and determine which kind of filter it is.

Many signals are not simply made up of one single frequency. They are a combination of frequencies, and this is where filters are useful. Try a wave function composed of two different frequency waves with the low-pass filter. Try this wave function with the high-pass filter.

Note

You can not change the frequency of this wave function with the slider bar in this animation.

8. What is the difference between the oscilloscope signals in the two cases?
9. Explain.

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Exploration 4: Phase Angle and Power

Assume an ideal power supply. The graph shows the voltage \(\color{red}{\text{(red)}}\) as a function of time across the source and the current \(\text{(black)}\) through the circuit (voltage is given in volts, current is given in milliamperes, and time is given in seconds).

Note

The initial current vs. time graphs are not centered about 0 because of the initial state of the capacitor and inductor.

 Restart.

To calculate the power dissipated by an RLC series circuit you cannot simply use \(I_{\text{rms}}V_{\text{rms}}\) because the power supply current and voltage are not in phase (unlike with a purely resistive load). This is due to the different phase shifts between voltage and current associated with the capacitors and inductors. The current through all elements must be the same, so the voltages across each are phase shifted from each other (see Illustrations 31.4 and 31.5). Here the equations for calculating power are

\[P = I_{\text{rms}}^{2}R = I_{\text{rms}}V_{\text{rms}}\cos\phi = I_{\text{rms}}^{2}Z\cos\phi\nonumber\]

where \(Z\) is the impedance of the series circuit (\(V_{\text{rms}}/I_{\text{rms}}\))and \(\phi\) is the phase shift between current and voltage defined as (\(\omega = 2\pi f\)):

\[Z = [R^{2}+(\omega L-1/\omega C)^{2}]^{1/2}\quad\text{and}\quad\cos\phi = R/Z \nonumber\]

1. Pick a frequency value. Find the impedance from \(V_{\text{rms}}/I_{\text{rms}}\).
2. Compare this value with the calculated value found using the equation above.
3. Calculate the phase shift.
4. Compare this calculated value with the value of phase shift measured directly from the graph. To measure the phase angle, since one period (\(1/f\)) represents a phase shift of \(2\pi\), measure the time difference between the peaks of the voltage and current plots and divide by the period (the time between the peaks of the voltage or the current) to find the percentage of \(2\pi\) by which the current is shifted.
5. What is the power dissipated?

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Exploration 5: RL Circuits and Phasors

Assume an ideal power supply. The graph shows the voltage as a function of time across the source \(\color{red}{\text{(red)}}\), the resistor \(\color{blue}{\text{(blue)}}\), and the inductor \(\color{green}{\text{(green)}}\) (voltage is given in volts and time is given in seconds). Restart.

To analyze the currents and voltages in this circuit, notice that you cannot simply use \(V = I X\) along with the peak current values. You must account for the phase differences between the voltages and the currents. One way to account for the phase differences is to describe the voltage, current, and reactance with phasors. Next to the circuit is an animation that shows the phasor representation of the circuit elements (this allows us to show the phase difference) in which the inductor voltage \(\color{green}{\text{(green)}}\) is \(\pi /2\) ahead of the resistor voltage \(\color{blue}{\text{(blue)}}\). The magnitude of each vector is the peak voltage across the element.

1. Since there is no phase shift between the resistor current and the voltage across the resistor, what would a phasor for the current look like in the phasor diagram?
2. Why is the phasor for the inductor \(\pi /2\) ahead of the resistor voltage (i.e., the current through the circuit)? (Hint: Does the inductor current lead or lag the voltage [see Illustration 31.4]?)
3. How does the length of the inductor's phasor change as you change the frequency?
4. Why does the length change as a function of frequency?

Notice that the phasors rotate at an angular speed of \(\omega =2\pi f\). The projection of any given voltage phasor on the \(y\) axis is the voltage across the circuit element at that time.

5. Pause the animation and explain how you can tell that the phasor diagram matches the voltages across the different circuit elements shown in the graph. In other words, verify that the \(y\) component of the vector in the phasor animation matches the value of the voltage shown on the graph.

From the vectors on the phasor diagram, we can develop a connection between the peak (or rms) voltage and the peak (or rms) current, where \(V_{0} = I_{0}Z\), and the phase difference between the voltage and current is given by \(\phi\). On the phasor diagram \(V_{0}\) (\(\color{red}{\text{the source voltage-red}}\)) is the vector sum of the two voltage vectors (\(\color{blue}{\text{resistor-blue}}\) and \(\color{green}{\text{inductor-green}}\)), and \(\phi\) is the angle between V₀ and the current (in the same direction as the resistor voltage phasor on the diagram). Exploration 31.6 develops the use of phasors for RLC circuits.

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Exploration 6: RLC Circuits and Phasors

Assume an ideal power supply. The graph shows the voltage as a function of time across the source \(\color{red}{\text{(red)}}\), the resistor \(\color{blue}{\text{(blue)}}\), the capacitor \(\color{green}{\text{(green)}}\), and the inductor \(\color{yellow}{\text{(yellow)}}\), as well as the current through the circuit (black) (voltage is given in volts, current is given in milliamperes, angles are given in degrees, and time is given in seconds). Restart.

From the vectors on the phasor diagram, we can develop a connection between the peak (or rms) voltage and the peak (or rms) current, where \(V_{0}=I_{0}Z\) and the phase difference between the voltage and current is given by \(\phi\). On the phasor diagram \(V_{0}\) (\(\color{red}{\text{the source voltage-red}}\)) is the vector sum of the three voltage vectors (\(\color{blue}{\text{resistor-blue}}\), \(\color{yellow}{\text{inductor-yellow}}\), and \(\color{green}{\text{capacitor-green}}\)) and \(\phi\) is the angle between \(V_{0}\) and the resistor phasor (since resistor current and voltage are in phase). The y components of the phasor vectors are the voltages across the various circuit elements. See Illustrations 31.6 and 31.7 as well as Exploration 31.5.

1. Explain the phase difference between the blue, yellow, and green vectors in the phasor animation.
2. Pick a frequency and pause the animation. Verify that the red vector is the vector sum of the other three vectors.
3. Pick a frequency and measure \(\phi\) on the phasor diagram using the pink protractor.
4. Explain how you can tell that the phasor animation matches the voltage and current vs. time graph for the circuit.
5. Also measure the phase angle on the voltage and current graphs. To measure the phase angle, since one period (\(1/f\)) represents a phase shift of \(2\pi\), measure the time difference between the peaks of the voltage and current plots and divide by the period.
6. Measure \(Z\) for this same frequency (\(Z = V_{0}/I_{0}\)).
7. Check your answers by using the equations for impedance and the phase shift between the voltage and the current, \(Z = (R^{2} + (\omega L - 1/\omega C)^{2})^{1/2}\) and \(\cos\phi = R/Z\).

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Exploration 7: RLC Circuit

Assume ideal components. The graph shows the voltage across the source \(\color{red}{\text{(red)}}\) and the current from the source \(\text{(black)}\) as functions of time (voltage is given in volts, current is given in milliamperes, and time is given in seconds). Restart.

An RLC circuit is similar to an oscillating spring or child on a swing. If you push the swing at exactly the same frequency as the natural frequency of oscillation (the most common way to push a swing), it quickly goes higher and higher. But if you push (or pull) part way through the swing at times that do not match the natural rhythm of the swing, the swing will not go as high as quickly and might even swing lower (in a fairly jerky fashion). When the current is the largest, this is called the resonance.

1. What is the resonant frequency of this circuit?
2. As you move the frequency of the driving source closer to the natural frequency of oscillation, what happens to the voltage and current?
3. Pick a new value for the variable resistor. What is the resonant frequency of this circuit?
4. What are the differences in the resonances with different values of \(R\)?
5. Compare the resonant frequency to \((1/2\pi )(1/LC)^{1/2}\). It should be the same.

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

Exploration 8: Damped RLC

Assume ideal components. The graph shows the voltage across the capacitor \(\color{red}{\text{(red)}}\), the voltage across the inductor \(\color{blue}{\text{(blue)}}\), and the voltage across the resistor \(\color{green}{\text{(green)}}\) as functions of time (voltage is given in volts).

Change the switches as you explore the behavior of this circuit. Restart.

1. Pick a specific time, measure the voltages on the graph, and verify that Kirchhoff's law holds when the switch is open and when the switch is closed.
2. What determines the time between peaks of the voltage when you close the switch?
3. Change the value of the variable resistor. What happens to the time for the oscillations if the resistor is large? When the resistor is small? Explain.

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian and Anne J. Cox.

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