Experiment 4

AC Circuits

4.1 Purpose

You will be investigating the voltage and current response of an alternating circuit (AC) to various impedance changes. The experiment will make extensive use of the digital oscilloscope.

4.2 Introduction

An alternating current, or AC, circuit is a device containing the usual elements (resistors, capacitors, etc.), but driven by a voltage that varies in time. In this experiment, we will study the behavior of AC circuits using series combinations of a resistor R R RRR, inductor L L LLL, and capacitor C C CCC, as depicted in Fig. 4.1.
Figure 4.1: A series R L C R L C RLCR L CRLC circuit driven by a sinusoidal voltage, with components voltages indicated.
circuit's behavior by writing down Kirchhoff's Loop Rule:
∑ Δ V = 0 = V ( t ) − V R − V L − V C V ( t ) = I R + L d I d t + q C = I R + L d I d t + 1 C ∫ I d t ∑ Δ V = 0 = V ( t ) − V R − V L − V C V ( t ) = I R + L d I d t + q C = I R + L d I d t + 1 C ∫ I d t {:[sum Delta V=0=V(t)-V_(R)-V_(L)-V_(C)],[V(t)=IR+L(dI)/(dt)+(q)/(C)],[=IR+L(dI)/(dt)+(1)/(C)int Idt]:}\begin{aligned} \sum \Delta V & =0=V(t)-V_{R}-V_{L}-V_{C} \\ V(t) & =I R+L \frac{d I}{d t}+\frac{q}{C} \\ & =I R+L \frac{d I}{d t}+\frac{1}{C} \int I d t \end{aligned}∑ΔV=0=V(t)−VR−VL−VCV(t)=IR+LdIdt+qC=IR+LdIdt+1C∫Idt
We would like to solve this differential equation for I ( t ) I ( t ) I(t)I(t)I(t) - and hence V R , V L V R , V L V_(R),V_(L)V_{R}, V_{L}VR,VL, and V C V C V_(C)V_{C}VC — given a driving voltage V ( t ) V ( t ) V(t)V(t)V(t). For arbitrary V ( t ) V ( t ) V(t)V(t)V(t), the equation is quite difficult to analyze. However, if V ( t ) V ( t ) V(t)V(t)V(t) is sinusoidal, driving the circuit with an angular frequency ω = 2 π f = 2 π / T ω = 2 π f = 2 π / T omega=2pi f=2pi//T\omega=2 \pi f=2 \pi / Tω=2πf=2π/T such that
V ( t ) = V max sin ⁡ ω t + ϕ V I ( t ) = I max sin ⁡ ω t + ϕ I , V ( t ) = V max sin ⁡ ω t + ϕ V I ( t ) = I max sin ⁡ ω t + ϕ I , {:[V(t)=V_(max)sin omega t+phi_(V)],[I(t)=I_(max)sin omega t+phi_(I)","]:}\begin{aligned} V(t) & =V_{\max } \sin \omega t+\phi_{V} \\ I(t) & =I_{\max } \sin \omega t+\phi_{I}, \end{aligned}V(t)=Vmaxsin⁡ωt+ϕVI(t)=Imaxsin⁡ωt+ϕI,
then the current in each circuit element will also be sinusoidal, though perhaps out of phase with the voltage across that element.

Voltage and Current in the Resistor R R RRR

By Ohm's Law, the voltage across the resistor is V R = I R V R = I R V_(R)=IRV_{R}=I RVR=IR. Because the resistor is in series with the voltage supply, the current running through it must have the same amplitude and phase as I ( t ) I ( t ) I(t)I(t)I(t). Substituting I ( t ) I ( t ) I(t)I(t)I(t) into Ohm's Law, we find
V R = I ( t ) ⋅ R = I max R sin ⁡ ω t V R = I ( t ) ⋅ R = I max R sin ⁡ ω t V_(R)=I(t)*R=I_(max)R sin omega tV_{R}=I(t) \cdot R=I_{\max } R \sin \omega tVR=I(t)⋅R=ImaxRsin⁡ωt
That is, the voltage and current in the resistor have the same time dependence, and are in phase with each other.

Voltage and Current in the Inductor L L LLL

For a circuit element like the inductor, the voltage can be obtained from Faraday's Law,
V L = L d I d t V L = L d I d t V_(L)=L(dI)/(dt)V_{L}=L \frac{d I}{d t}VL=LdIdt
Again, because the inductor is in series with the voltage supply and the resistor, the current moving through it is the same I ( t ) I ( t ) I(t)I(t)I(t) used above. Substituting and differentiating I ( t ) I ( t ) I(t)I(t)I(t), we obtain
V L = ω L I max cos ⁡ ω t = ω L I max sin ⁡ ( ω t + π 2 ) = I max X L sin ⁡ ( ω t + π 2 ) V L = ω L I max cos ⁡ ω t = ω L I max sin ⁡ ω t + π 2 = I max X L sin ⁡ ω t + π 2 {:[V_(L)=omega LI_(max)cos omega t],[=omega LI_(max)sin(omega t+(pi)/(2))],[=I_(max)X_(L)sin(omega t+(pi)/(2))]:}\begin{aligned} V_{L} & =\omega L I_{\max } \cos \omega t \\ & =\omega L I_{\max } \sin \left(\omega t+\frac{\pi}{2}\right) \\ & =I_{\max } X_{L} \sin \left(\omega t+\frac{\pi}{2}\right) \end{aligned}VL=ωLImaxcos⁡ωt=ωLImaxsin⁡(ωt+π2)=ImaxXLsin⁡(ωt+π2)
The quantity X L = ω L X L = ω L X_(L)=omega LX_{L}=\omega LXL=ωL is called the reactance of the inductor. The comparison of X L X L X_(L)X_{L}XL to the resistance R R RRR in Ohm's Law is actually a good way to think about how the inductor works. For an inductor, the current running through the device will be out of phase with the voltage across it.
There are two things you should note about this result:
  1. In the inductor, the voltage and current are out of phase; the voltage "leads" the current by a phase shift ϕ = + 90 ∘ ϕ = + 90 ∘ phi=+90^(@)\phi=+90^{\circ}ϕ=+90∘.
  2. The maximum voltage occurs when sin ⁡ ( ω t + π / 2 ) = 1 sin ⁡ ( ω t + π / 2 ) = 1 sin(omega t+pi//2)=1\sin (\omega t+\pi / 2)=1sin⁡(ωt+π/2)=1; at this location,
V max = ω L I max V max = ω L I max V_(max)=omega LI_(max)V_{\max }=\omega L I_{\max }Vmax=ωLImax
By replacing ω L ω L omega L\omega LωL by the variable X L X L X_(L)X_{L}XL, we get an expression that looks rather like Ohm's Law:
V max = I max X L V max = I max X L V_(max)=I_(max)X_(L)V_{\max }=I_{\max } X_{L}Vmax=ImaxXL

Voltage and Current in the Capacitor C C CCC

For the capacitor, the voltage V C V C V_(C)V_{C}VC is obtained from the relationship
V C = q C = 1 C ∫ I ( t ) d t V L = 1 C ( − I max ω cos ⁡ ω t ) = 1 ω C I max sin ⁡ ( ω t − π 2 ) = X C I max sin ⁡ ( ω t − π 2 ) V C = q C = 1 C ∫ I ( t ) d t V L = 1 C − I max ω cos ⁡ ω t = 1 ω C I max sin ⁡ ω t − π 2 = X C I max sin ⁡ ω t − π 2 {:[V_(C)=(q)/(C)=(1)/(C)int I(t)dt],[V_(L)=(1)/(C)(-(I_(max))/(omega)cos omega t)],[=(1)/(omega C)I_(max)sin(omega t-(pi)/(2))],[=X_(C)I_(max)sin(omega t-(pi)/(2))]:}\begin{aligned} V_{C} & =\frac{q}{C}=\frac{1}{C} \int I(t) d t \\ V_{L} & =\frac{1}{C}\left(-\frac{I_{\max }}{\omega} \cos \omega t\right) \\ & =\frac{1}{\omega C} I_{\max } \sin \left(\omega t-\frac{\pi}{2}\right) \\ & =X_{C} I_{\max } \sin \left(\omega t-\frac{\pi}{2}\right) \end{aligned}VC=qC=1C∫I(t)dtVL=1C(−Imaxωcos⁡ωt)=1ωCImaxsin⁡(ωt−π2)=XCImaxsin⁡(ωt−π2)
The expression X C = 1 / ω C X C = 1 / ω C X_(C)=1//omega CX_{C}=1 / \omega CXC=1/ωC is also called the reactance; but unlike X L X L X_(L)X_{L}XL, it actually decreases as a function of ω ω omega\omegaω. The capacitor is like a mirror image of the inductor, strongly resisting the flow of current when the driving frequency ω ω omega\omegaω is small, but barely resisting the flow of high frequency signals.
As before, notice two things about this result:
  1. The voltage and current are out of phase, but the voltage now "lags" behind the current by a phase shift ϕ = − 90 ∘ ϕ = − 90 ∘ phi=-90^(@)\phi=-90^{\circ}ϕ=−90∘.
  2. At the maximum voltage, where sin ⁡ ( ω t − π / 2 ) = 1 sin ⁡ ( ω t − π / 2 ) = 1 sin(omega t-pi//2)=1\sin (\omega t-\pi / 2)=1sin⁡(ωt−π/2)=1,
V max = I max ω C V max = I max ω C V_(max)=(I_(max))/(omega C)V_{\max }=\frac{I_{\max }}{\omega C}Vmax=ImaxωC
Again, replacing 1 / ω C 1 / ω C 1//omega C1 / \omega C1/ωC by a new variable X C X C X_(C)X_{C}XC yields an equation that looks like Ohm's Law:
V max = I max X C V max = I max X C V_(max)=I_(max)X_(C)V_{\max }=I_{\max } X_{C}Vmax=ImaxXC
Figure 4.2: Summary of the various phase relationships for a series R L C R L C RLCR L CRLC circuit.

Impedance and Phase of the R L C R L C RLCR L CRLC Circuit

Now that we understand the voltages across each individual element V R , V C V R , V C V_(R),V_(C)V_{R}, V_{C}VR,VC, and V L V L V_(L)V_{L}VL, we can try to make sense of the overall R L C R L C RLCR L CRLC combination. The total voltage V R L C V R L C V_(RLC)V_{R L C}VRLC across the series combination is just the algebraic sum of the voltage drops across each component:
V R L C = V R + V L + V C = I max [ R sin ⁡ ω t + X L sin ⁡ ( ω t + π 2 ) + X C sin ⁡ ( ω t − π 2 ) ] (4.1) = I max [ R sin ⁡ ω t + ( X L − X C ) cos ⁡ ω t ] V R L C = V R + V L + V C = I max R sin ⁡ ω t + X L sin ⁡ ω t + π 2 + X C sin ⁡ ω t − π 2 (4.1) = I max R sin ⁡ ω t + X L − X C cos ⁡ ω t {:[V_(RLC)=V_(R)+V_(L)+V_(C)],[=I_(max)[R sin omega t+X_(L)sin(omega t+(pi)/(2))+X_(C)sin(omega t-(pi)/(2))]],[(4.1)=I_(max)[R sin omega t+(X_(L)-X_(C))cos omega t]]:}\begin{align*} V_{R L C} & =V_{R}+V_{L}+V_{C} \\ & =I_{\max }\left[R \sin \omega t+X_{L} \sin \left(\omega t+\frac{\pi}{2}\right)+X_{C} \sin \left(\omega t-\frac{\pi}{2}\right)\right] \\ & =I_{\max }\left[R \sin \omega t+\left(X_{L}-X_{C}\right) \cos \omega t\right] \tag{4.1} \end{align*}VRLC=VR+VL+VC=Imax[Rsin⁡ωt+XLsin⁡(ωt+π2)+XCsin⁡(ωt−π2)](4.1)=Imax[Rsin⁡ωt+(XL−XC)cos⁡ωt]
Now, we know that the solution V R L C V R L C V_(RLC)V_{R L C}VRLC must be sinusoidal, and must have some phase ϕ ϕ phi\phiϕ :
V R L C = V max sin ⁡ ( ω t + ϕ ) (4.2) = V max [ sin ⁡ ( ω t ) cos ⁡ ϕ + cos ⁡ ( ω t ) sin ⁡ ϕ ] V R L C = V max sin ⁡ ( ω t + ϕ ) (4.2) = V max [ sin ⁡ ( ω t ) cos ⁡ ϕ + cos ⁡ ( ω t ) sin ⁡ ϕ ] {:[V_(RLC)=V_(max)sin(omega t+phi)],[(4.2)=V_(max)[sin(omega t)cos phi+cos(omega t)sin phi]]:}\begin{align*} V_{R L C} & =V_{\max } \sin (\omega t+\phi) \\ & =V_{\max }[\sin (\omega t) \cos \phi+\cos (\omega t) \sin \phi] \tag{4.2} \end{align*}VRLC=Vmaxsin⁡(ωt+ϕ)(4.2)=Vmax[sin⁡(ωt)cos⁡ϕ+cos⁡(ωt)sin⁡ϕ]
Since eq. (4.1) and eq. (4.2) must be equal at all times t t ttt, we can equate the coefficients of sin ⁡ ω t sin ⁡ ω t sin omega t\sin \omega tsin⁡ωt and cos ⁡ ω t cos ⁡ ω t cos omega t\cos \omega tcos⁡ωt and solve for ϕ ϕ phi\phiϕ :
V max cos ⁡ ϕ = I max R V max sin ⁡ ϕ = I max ( X L − X C ) } ⇒ tan ⁡ ϕ = X L − X C R = ω L − 1 / ω C R V max cos ⁡ ϕ = I max R V max sin ⁡ ϕ = I max X L − X C ⇒ tan ⁡ ϕ = X L − X C R = ω L − 1 / ω C R {:[V_(max)^(cos phi),=I_(max)R],[V_(max)sin phi,=I_(max)(X_(L)-X_(C))]}=>tan phi=(X_(L)-X_(C))/(R)=(omega L-1//omega C)/(R)\left.\begin{array}{rl} V_{\max }^{\cos \phi} & =I_{\max } R \\ V_{\max } \sin \phi & =I_{\max }\left(X_{L}-X_{C}\right) \end{array}\right\} \Rightarrow \tan \phi=\frac{X_{L}-X_{C}}{R}=\frac{\omega L-1 / \omega C}{R}Vmaxcos⁡ϕ=ImaxRVmaxsin⁡ϕ=Imax(XL−XC)}⇒tan⁡ϕ=XL−XCR=ωL−1/ωCR
The overall phase shift of the R L C R L C RLCR L CRLC combination depends on the reactive parts of the circuit X L X L X_(L)X_{L}XL and X C X C X_(C)X_{C}XC. Since these are functions of frequency, the phase will change if the driving frequency ω ω omega\omegaω changes. As a final step, we can eliminate ϕ ϕ phi\phiϕ from these two equations by squaring and adding them:
V max = I max R 2 + ( X L − X C ) 2 = I max Z V max = I max R 2 + X L − X C 2 = I max Z {:[V_(max)=I_(max)sqrt(R^(2)+(X_(L)-X_(C))^(2))],[=I_(max)Z]:}\begin{aligned} V_{\max } & =I_{\max } \sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ & =I_{\max } Z \end{aligned}Vmax=ImaxR2+(XL−XC)2=ImaxZ
where Z = R 2 + ( X L − X C ) 2 Z = R 2 + X L − X C 2 Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}Z=R2+(XL−XC)2. Once again, this looks suggestively like Ohm's Law: voltage equals current times "resistance." The quantity Z Z ZZZ, called the impedance of the R L C R L C RLCR L CRLC circuit, is in fact just a generalized, frequency-dependent resistance. The resistive part of Z Z ZZZ (just the resistor R R RRR ) resists all signals in a frequency-independent way; it is also responsible for dissipating the power delivered by the voltage supply as heat. The reactive part of Z Z ZZZ (the combination X L − X C X L − X C X_(L)-X_(C)X_{L}-X_{C}XL−XC ) does not dissipate energy, but is responsible for determining the circuit's frequency response.
A way of visualizing this is using phasor diagrams. Notice how all the equations derived in this subsection can be obtained via Pythagoras Theorem from Fig. 4.3.
Figure 4.3: A series of phasor diagrams providing visual illustration on how the reactances influence the voltage and current of a RLC circuit. The voltage contributions of the components have to be treated as vector sums due to the phase differences between them.

Frequency Response of the R L C R L C RLCR L CRLC Circuit

Because the R L C R L C RLCR L CRLC circuit contains reactive elements, the voltage V V VVV across the resistor depends on the frequency ω ω omega\omegaω of the AC source. The frequency dependence, depicted in Fig. 4.4, has a resonance; that is, there is a frequency ω 0 ω 0 omega_(0)\omega_{0}ω0 where the impedance Z Z ZZZ takes on a minimum value, so the current takes on a maximum value for a given driving voltage.
Figure 4.4: Frequency response of the R L C R L C RLCR L CRLC circuit, showing the resonance frequency ω 0 ω 0 omega_(0)\omega_{0}ω0.
The impedance is clearly a minimum when X L = X C X L = X C X_(L)=X_(C)X_{L}=X_{C}XL=XC, so that the resonance frequency is
ω 0 L = 1 ω 0 C ω 0 = 1 L C ω 0 L = 1 ω 0 C ω 0 = 1 L C {:[omega_(0)L=(1)/(omega_(0)C)],[omega_(0)=(1)/(sqrt(LC))]:}\begin{aligned} \omega_{0} L & =\frac{1}{\omega_{0} C} \\ \omega_{0} & =\frac{1}{\sqrt{L C}} \end{aligned}ω0L=1ω0Cω0=1LC
The resonant behavior of the R L C R L C RLCR L CRLC circuit actually has a pretty intuitive explanation. Remember that at low frequencies, the capacitor has a high "resistance" to current flow, but the inductor allows current to move freely. At high frequencies, it is the inductor that resists the signal, while the capacitor has a very low resistance. In between, there is a crossover region around ω 0 ω 0 omega_(0)\omega_{0}ω0 where the influences of both effects are minimal.
The width of the peak at half of the maximum voltage or current, denoted Δ ω Δ ω Delta omega\Delta \omegaΔω or FWHM (full width at half maximum) depends on the size of the resistor in the R L C R L C RLCR L CRLC combination. Smaller resistors will produce sharper peaks, making the circuit more sensitive to frequencies near ω 0 ω 0 omega_(0)\omega_{0}ω0.
The frequency response of the R L C R L C RLCR L CRLC combination makes it a very useful filter for AC signals. Called a bandpass filter, circuits like this are employed when we want to pick out a frequency from a range of possible signals. A perfect example is the radio tuner in an automobile.

4.3 Experiment

You will record several observations of the behavior of the series R L C R L C RLCR L CRLC circuit shown in Fig. 4.5. The procedure makes heavy use of the digital oscilloscope used in the Capacitance experiment. If you still feel uncomfortable using the scope, refer back to that section of the lab manual to refresh your skills.
Figure 4.5: Setup of the R L C R L C RLCR L CRLC circuit, function generator, and scope.
Besides the oscilloscope, there are only a few pieces of equipment you will need to use:
  1. A variable resistor, in the form of two resistors in series ( 1.2 K Ω 1.2 K Ω 1.2 K Omega1.2 K \Omega1.2KΩ and 3.3 K Ω 3.3 K Ω 3.3 K Omega3.3 K \Omega3.3KΩ ).
  2. A capacitor provided for you in the lab. Be sure to note the capacitance.
  3. Two inductors: a small 150 mH circuit component, and a large induction ring of unknown L L LLL.
  4. A function generator, which you will use to send sinusoidal signals through the circuit.

4.4 Procedure

Setting Up

The first thing you should do is look at your driving signal on the oscilloscope.
  • Connect the output of the function generator to CH 1 on the oscilloscope using wire leads of a single color.
  • Make sure the ground on the function generator is connected to the ground on the scope. Ground is indicated GND on the black female BNC-to-banana plug connector.
  • Set up the function generator to produce a sinusoidal signal in the range of 1 kHz , with a peak-to-peak voltage of V p p = 20 V V p p = 20 V V_(pp)=20VV_{p p}=20 \mathrm{~V}Vpp=20 V.
  • Set up the oscilloscope to trigger on the function generator. You can do this by pressing the Trigger button, and then on the menu that appears on the right-hand side of the display, press the button next to trigger source until CH 1 appears.
You should now be able to alter the VOLTS/DIV and SEC/DIV settings in order to view the sine wave produced by the function generator. If you want to take a shortcut, you can quickly obtain the
appropriate settings by pressing the AUTOSET button. Note also the following general hints for dealing with the equipment:
  • If at any time the scope display is very noisy, it is possible that you have lost the trigger. You can fix this by adjusting the trigger level knob in the TRIGGER controls, or turn off CH 2 and press AUTOSET.
  • The frequency recorded by the scope reads is f f fff, in Hz , or cycles per second. The angular frequency of your R L C R L C RLCR L CRLC circuit, ω ω omega\omegaω, given in rad sec − 1 sec − 1 sec^(-1)\sec ^{-1}sec−1, is related to f f fff by ω = 2 π f ω = 2 π f omega=2pi f\omega=2 \pi fω=2πf.

Resonance

  • Use a multimeter to measure the resistance of the 150 mH inductor. Connect the inductor, the capacitor, and the resistors in series with the function generator using cables of a single color (different from that used above), as depicted in Fig. 4.5.
  • Connect CH 2 of the oscilloscope across the resistor, making sure that the ground on the oscilloscope and the ground side of the resistor are connected to the ground on the function generator; use wires of a color different than the two colors used in earlier procedure steps. Your oscilloscope and function generator now share a common ground.
  • Set up the circuit so you are using a resistance of 3.3 K Ω 3.3 K Ω 3.3KOmega3.3 \mathrm{~K} \Omega3.3 KΩ. Make sure you see a stable display of both CH 1 and CH 2.
  • Using your circuit components, estimate the resonance frequency you expect to observe.
  • Use the frequency knob on the function generator to try to find the circuit resonance. Sweep through a set of frequencies until you observe the signal on CH 2 peaking around some value. NOTE: observe when the peak-to-peak voltage of the sine wave on CH 2 is maximized.
  • When you have found the resonance frequency ω 0 ω 0 omega_(0)\omega_{0}ω0, slowly scan over frequencies and record the peak-to-peak voltage you observe in CH 2. You can use the scope's MEASURE function to automatically display the peak-to-peak signal amplitude.
  • Make sure you scan through a wide enough frequency range that you can record the FWHM of the signal. Record the voltage amplitude at 20 points (at least) around the location of the resonance.
Repeat this procedure for a lower value of the resistance ( 1.2 K Ω 1.2 K Ω 1.2KOmega1.2 \mathrm{~K} \Omega1.2 KΩ ), and a higher value (4.5K Ω Ω Omega\OmegaΩ ). Make sure you record the value of R R RRR for each voltage table you record.
When you are finished, replace the 150 mH inductor with the large copper ring. Be careful: the ring is quite heavy! Sweep with the function generator through a range of frequencies until you find the
resonance frequency ω 0 ω 0 omega_(0)\omega_{0}ω0. Using this measurement and the value of C C CCC, estimate (with uncertainties) the inductance L L LLL of the copper ring.

Phase of Driving Voltage and V R V R V_(R)V_{R}VR

  • Replace the copper ring in your setup with the small inductor you used earlier.
  • Set up your circuit with the resistor of 3.3 K Ω 3.3 K Ω 3.3 K Omega3.3 K \Omega3.3KΩ.
  • Make sure you can see the driving signal on CH 1 and the voltage across the resistor on CH 2 . Note whether the voltage across the resistor is ahead of, behind, or the same as the driving voltage.
  • Vary the frequency of the function generator to determine the relationship between the driving voltage and the current in the circuit at, above and below the resonant frequency ω 0 ω 0 omega_(0)\omega_{0}ω0, and record your qualitative observations.
  • For five values of the frequency at and around ω 0 ω 0 omega_(0)\omega_{0}ω0, measure the phase difference between the driving voltage V ( t ) V ( t ) V(t)V(t)V(t) and the current in the resistor. You can do this using the CURSOR function of the scope. Measure the total time for one full cycle of the driving signal, T d T d T_(d)T_{d}Td, and then measure the time difference t R − t d t R − t d t_(R)-t_(d)t_{R}-t_{d}tR−td between the max max max\maxmax (or min min min\minmin ) of the driving signal and the max (or min min min\minmin ) of the signal across the resistor.
  • Be sure to record error and propagate it. The phase difference ϕ ϕ phi\phiϕ is equal to
ϕ = 2 π t R − t d T d ϕ = 2 π t R − t d T d phi=2pi(t_(R)-t_(d))/(T_(d))\phi=2 \pi \frac{t_{R}-t_{d}}{T_{d}}ϕ=2πtR−tdTd
Note that this formula implies that a positive phase shift corresponds to the driving voltage peaking earlier, and therefore to the left of, the resistor voltage.
Phase of V ( t ) , V L V ( t ) , V L V(t),V_(L)V(t), V_{L}V(t),VL, and V C V C V_(C)V_{C}VC
Above you found that the driving voltage and the voltage across the resistor (and thus the current in the circuit) are in phase at the resonant frequency. In this last part of the lab you will observe the phase differences between the driving signal V ( t ) V ( t ) V(t)V(t)V(t), the inductor V L V L V_(L)V_{L}VL, and the capacitor V C V C V_(C)V_{C}VC.

Phase Shift of the Inductor

Keep the circuit in its original configuration, as in Fig. 4.5, but increase the frequency of the function generator to a value well above resonance - e.g., 10 kHz . At a sufficiently high frequency, X L = X L = X_(L)=X_{L}=XL= ω L → ω L → omega L rarr\omega L \rightarrowωL→ big, and X C = 1 / ω C → X C = 1 / ω C → X_(C)=1//omega C rarrX_{C}=1 / \omega C \rightarrowXC=1/ωC→ small. That is, the inductor dominates the circuit, so its reactance will determine the overall phase shift ϕ ϕ phi\phiϕ that you observe in the voltage across the resistor.
To determine the phase shift, just measure the phase of the voltage across the resistor as before. It is only necessary to make one measurement in this case. The expression for the measured phase shift also remains the same as before.

Phase Shift of the Capacitor

Again, leave the configuration of the circuit alone, but decrease the frequency to well below resonance - e.g., 10 to 50 Hz .
At this low frequency, X L = ω L → X L = ω L → X_(L)=omega L rarrX_{L}=\omega L \rightarrowXL=ωL→ small, and X C = 1 / ω C → X C = 1 / ω C → X_(C)=1//omega C rarrX_{C}=1 / \omega C \rightarrowXC=1/ωC→ big. That is, the capacitor dominates the circuit, and its reactance determines the overall phase shift ϕ ϕ phi\phiϕ that you observe in the voltage across the resistor. To determine the phase shift, the procedure is the same as what you tried above.

Summary of data:

  • Resonance section:
  • Capacitance C C CCC and inductance L L LLL of elements
  • V p p V p p V_(pp)V_{p p}Vpp vs frequency, for three values of resistance R R RRR
  • Resonance frequency ω 0 ω 0 omega_(0)\omega_{0}ω0, for unknown L L LLL
  • Phase shifts:
  • Phase difference between voltage and current: frequencies close to resonance
  • Phase shift for high frequency
  • Phase shift for low frequency

4.5 Analysis

Resonance of the RLC Circuit

  1. For the three resistor values you used, plot the peak-to-peak voltages you recorded as a function of frequency. Put all three plots on the same set of axes, and normalize each to the maximum voltage from each dataset.
  2. What is the resonant frequency of the circuit you observed? Report an angular frequency, ω ω omega\omegaω. Be sure to record an error for your measurement.
  3. Calculate what the resonant frequency should in theory be by reading the values off the circuit elements and using the formula for at the resonant frequency. Remember to propagate errors. Does your measured value agree with the theoretical value within error? What is your relative accuracy?
  4. Using your plots, calculate the FWHM Δ ω Δ ω Delta omega\Delta \omegaΔω for each set of data. Be sure to estimate an uncertainty.
  5. What happens to the resonant frequency when R R RRR is increased? What happens to the full width at half max?
  6. For the L L LLL you measured using the large copper ring, how does the estimate compare to the value marked on the inductor?

Phase of Driving Voltage and V R V R V_(R)V_{R}VR

Compare your measured values of the phase difference with the theoretical value derived for ϕ ϕ phi\phiϕ :
tan ⁡ ϕ = ω L − 1 / ω C R tan ⁡ ϕ = ω L − 1 / ω C R tan phi=(omega L-1//omega C)/(R)\tan \phi=\frac{\omega L-1 / \omega C}{R}tan⁡ϕ=ωL−1/ωCR
Remember to use the value for f f fff recorded from the oscilloscope and convert this to an angular frequency ω ω omega\omegaω. Assume that the internal resistances of the circuit components, other than the chosen resistance value from the resistors in series, are negligible. Does your measured value agree with the theoretical value, within error? If there is a large discrepancy, does accounting for the 50 Ω 50 Ω 50 Omega50 \Omega50Ω output impedance of the function generator help?

Phase Difference of V ( t ) V ( t ) V(t)V(t)V(t) and the Capacitor/Inductor

  • Does the inductor voltage V L V L V_(L)V_{L}VL lead the driving voltage by 90 ∘ 90 ∘ 90^(@)90^{\circ}90∘, as expected?
  • Does the capacitor voltage V C V C V_(C)V_{C}VC lag by 90 ∘ 90 ∘ 90^(@)90^{\circ}90∘ ?
  • If there are any discrepancies, can they be explained by errors in your measurements?

General Questions

  • We mentioned that the inductor resists signals with high frequencies. What property of the inductor causes it to behave this way? HINT: Think of an inductor as a current loop or solenoid, and consider what happens when you try to vary the current through the loop.
  • Similarly, given what you know about how capacitors charge and discharge, why do you think it tends to block DC signals but likes high frequency AC signals?
  • You observed the behavior of the capacitor and inductor indirectly, by looking at the signal across the resistor. Why is this necessary?