Block C - Sinusoidal Response

Objectives

Measure AC voltages and currents in RC, RL, and RLC AC networks and sketch phasor diagrams.

Background

In Block B, we examined the natural response of various circuits. In this experiment, we want to consider the forced response of the same simple circuits. The forcing function will be a 60 Hz sinusoid, so we can also call it the sinusoidal response.

One of the benefits of considering the sinusoidal response of circuits is that the analysis can be done by algebra rather than by differential equations. This decreases the effort involved and also provides additional insights into the behavior of circuits.

Consider the series RLC circuit in Fig. 1. V is a sinusoidal voltage source described by its rms amplitude and a phase angle. We usually consider the source to be the reference so that its angle is 0 degrees. The inductor and capacitor have reactances given by

X_L = L, X_C = 1/C .........(1)

where = 2 f. The corresponding impedances are

where j = , an imaginary number.

The total impedance of the circuit is

The magnitude of the impedance is

.........(4)

and its phase angle is

.........(5)

The current is given by

.........(6)

If X_L is greater than X_C, then is positive and the corresponding phase angle of I is negative. This situation is shown in Fig. 2 and Fig. 3. These figures contain a great deal of information, which we will try to explain in detail.

Both figures are plotted on the complex plane. In Fig. 2, R is on the positive real axis, jX_L is on the positive y axis (the positive imaginary axis), and -jX_C is on the negative imaginary axis. The complex number Z is formed geometrically by combining R and j(X_L - X_C) as shown. It will be in the first quadrant if is positive and in the fourth quadrant if is negative. If the two reactances are equal, then they cancel out and Z is equal to R. This condition is known as resonance.

It should be mentioned that the arrows used to represent complex numbers can be moved parallel to themselves as required to form desired triangles. This is different from force diagrams, say, where the location of a force arrow (or vector) has physical significance. If a force vector is moved parallel to itself, the torque or moment arm changes, and we no longer have the same problem. The complex plane is not the xy plane of statics and some of the rules are different.

The diagram of voltages and current corresponding to Fig. 2 is shown in Fig. 3. This is called a phasor diagram. The input voltage is taken as the reference so we put it on the positive real axis. The circuit is inductive (X_L > X_C) so the current is in the fourth quadrant. We say that the current lags the voltage in an inductive circuit. In an ideal inductor, the current lags the voltage by 90^o (or the voltage leads the current by 90^o).

The voltage V is composed of three separate voltages by Kirchhoff's voltage law.

V_R is in phase with I, so it lies along the same line. The voltage V_L leads I by 90^o , which is shown on the diagram by rotating 90^o counterclockwise. The voltage V_C lags I by 90^o, so we rotate its arrow clockwise by 90^o. V_R and (V_L + V_C) form the two sides of a right triangle, with V as the hypotenuse. The diagram therefore shows that Kirchhoff's voltage law is satisfied.

The diagram is drawn with V_L larger than V, which is quite possible in this resonant circuit. In fact, both V_L and V_C may be larger than V. This effect is used to increase the voltage received by a radio antenna at the frequency of the desired station, before the voltage is amplified by the circuitry in the radio.

Example. Let V = 120 < 0^o, R = 5 , X_L = 25 , and X_C = 10 . What is I, V_R, V_L, V_C, and the power dissipated in the circuit?

Z = 5 + j(25 - 10) = 5 + j15 = 15.8171.57^o ..........(8)

I = V/Z = 120/15.8171.57^o = 7.59-71.57^o ..........(9)

V_R = R I = 5(7.59-71.57^o) = 37.95-71.57^o ..........(10)

V_L = Z_L I = j X_L I = (25)(7.59)(90 - 71.57)^o = 189.7418.43^o ..........(11)

V_C = Z_C I = -j X_C I = (10)(7.59)(-90 - 71.57)^o = 75.89-161.57^o ..........(12)

Note that multiplying by j is the same as adding 90^o to the angle, while multiplying by -j is the same as subtracting 90^o from the angle. Also note that the power P is a real number. It never has any angle associated with it. We might think of the resistor not being able to know what voltage reference was chosen for the circuit, so that it always chooses its own voltage as reference. We could get the same result by P = |V_R|²/R = (37.95)²/5 = 288 W. The same result is obtained by P = |V_R| |I|. Be careful to only use the current through and the voltage across the resistor in calculating power. The voltages across the inductor and capacitor do not have anything to do with power.

You should practice the above calculations with your calculator, to make sure you are able to do them yourself.

In observing these quantities on the scope, we will see sinusoids with a phase shift between them rather than arrows, such as the sketch of v and v_R in Fig. 4. We know from the above calculations that V_R is lagging V, but how can we determine that by just looking at the scope? The answer is to think of time as increasing to the right, so the second peak must be lagging the first peak as we encounter the peaks in moving from left to right across the screen.

Discussion and Calculations

1. For the circuit in Fig. 5, theoretically determine I, V_R, and V_c if V = 6 0^o Volts and f = 60 Hz. Draw the phasor diagram of V, V_R, V_c, and I, reasonably close to scale. By what angle does V_c lead or lag V?

2. What time/div setting should you use on the horizontal scale of the scope to see at least one but not two full periods of this signal?

3. If the phase shift between two 60 Hz waveforms is 90 degrees, what is the time shift in milliseconds you should see on the scope?

Instructional Activity in Class

1. a. Assemble the circuit of Fig. 6a and observe the waveforms of I and V_c. Push the "2" button and select the Invert mode to make the scope display the negative of CH2. Sketch the waveforms in your notebook, noting peak amplitudes and positions of zero crossings on your sketch. Determine the phase of I with respect to V_c.

   b. Rearrange the circuit to that of Fig. 6b. Make sure you are no longer inverting CH2. Observe the waveforms of I and V. Sketch the waveforms and determine the phase of I with respect to V.

   c. Rearrange the circuit to that of Fig. 6c. Observe the waveforms of V_c and V. Sketch the waveforms and determine the phase of V_c with respect to V. How does your result compare with the result of Discussion and Calculations part1 ? In your writeup draw a phasor diagram of I, V_R, V_c, and V using values measured above and discuss results. Use rms phasors.

2. Locate the coil mounted between two plastic plates that you used in Block B. Measure the DC resistance of this coil using a digital multimeter. Then locate a one Ohm resistor mounted on a block of wood and stored under the bench. Assemble the circuit in Fig. 7. Note that the voltage across the one Ohm resistor is numerically equal to the current through it. Sketch the waveforms of V_coil and I in your notebook. Use = 377 and the magnitudes of V_coil, I, and R to compute L. Then use the measured angle between V_coil and I plus the measured value of R to compute another estimate of L. In your writeup draw the phasor diagram of I, V_coil, and V_R using values measured above. Why does I not lag V_coil by 90^o? Discuss.

3. Assemble the circuit in Fig. 8 and measure V, V_coil, V_c, and V_R with the digital multimeter. Measure the phase of I with respect to V by using the oscilloscope. Does I lead or lag V? Sketch the waveforms so you can verify your answer later. In your writeup draw the phasor diagram of I, V_R, V_c, V_coil, and V to scale. Assume I 0^o as the reference phasor. The magnitude of V_R and V_c are known experimentally and the phase angles of V_R and V_c with respect to I are known theoretically. Using the magnitudes of V_coil and V as measured, and the KVL (Kirchoff's Voltage Law) equation of the circuit, locate V_coil and V graphically and numerically to complete the phasor diagram of voltages. Compare the phase angle between V and I with that previously measured experimentally.

4. With the same circuit as in Lab Activity 3, insert an iron bar (obtained from the instructor) in the coil and observe the effect. Can you cause the circuit to go from capacitive to inductive as more iron is inserted? Can you make the circuit appear as a resistive circuit, with current in phase with the voltage?

5. Measure the real power absorbed by the 1 Ohm resistor and the real power absorbed by the circuit past the transformer. Clamp the current probe of the Fluke Harmonic Analyzer near the 1 Ohm resistor. Connect the negative voltage clamp (black) in the postion in Fig. 8 marked GND and move the postive clamp (red) to measure the appropriate voltage.

Conclusion

1. Go through the lab and make the calculations and draw the phasor diagrams requested.

2. In Lab Activity 2, you used two different techniques to determine an estimate of L. Which technique would be the most accurate, and why?

3. Calculate the real power (P in Watts) in the 1 Ohm resistor using the measurements of Lab Activity 3 and compare your results with your measurements in Lab Activity 5.

4. How much power is absorbed by the coil and capacitor in Fig. 8? Estimate the combined resistance of the coil and capacitor using the measurements made in this lab.
5. Discuss the other questions asked during the lab.