A potentiometer, a capacitor, one inductor and a signal generator are linked in series. Top top the oscilloscope you deserve to read the voltage across the signal generator, the potentiometer and either the inductor or the capacitor (connected come the oscilloscope by the double-pole, double-throw switch at right in the photograph). When the generator to produce a sine wave, you can see the phase connections of the voltages across the components, through respect to each other and to the voltage used to the circuit. Once you turn the dial the the signal generator with the resonant frequency (in one of two people direction), all the voltages (except that across the signal generator, that course) rise to a maximum, then decrease thereafter. The phase connections also change as presented below. In addition, girlfriend can discover the impact of damping top top the “quality,” or “Q,” the the circuit through adjusting the potentiometer (*vide infra*). * Important note: * The potentiometer in this circuit is large enough the you can take the mechanism to critical damping, and also then you can overdamp it. (See show 72.66 -- decaying oscillations in an LRC circuit.) Under these problems the resonance becomes broad, and also the change in frequency crucial to do either the capacitive reactance or the inductive reactance substantially greater 보다 the resistance, is bigger than for the underdamped situation. Also, the present maximum, and thus the voltage across each the the components, becomes smaller (

*vide infra*).

You are watching: In an lrc circuit, when the inductive and capacitive reactances are equal,

The schematic over shows the circuit (circuit A). One overhead transparency of this schematic is easily accessible for girlfriend to display in class, or you may download a .pdf version here to show on the data projector. Channel 3 on the oscilloscope shows the voltage used by the signal generator to the circuit, and channel 4 shows the voltage across the resistor. Channels one and also two are associated via a double-pole, double-throw switch to the end of either the capacitor (switch down) or the inductor (switch up). The oscilloscope subtracts the signal ~ above channel two from the on channel one, and displays the distinction on a separate trace.

You can also have the circuit wired to display the voltages throughout the capacitor and the inductor simultaneously, and also to switch in between the capacitor and also the resistor (circuit B):

An overhead transparency of this circuit is likewise available, or you can download the .pdf document here. Here, oscilloscope channel 4 reads the potential throughout the inductor, and channels one and two walk to one of two people the capacitor or the resistor via the double-pole, double-throw switch. Traces acquired with both circuits appear below.

**Please note, as soon as you inquiry this demonstration, which circuit (A or B) you would like. While either circuit allows you to show in information the behavior of one LRC circuit, depending on which details things friend would choose to have the ability to show simultaneously, you might prefer one over the other.**

The potential across the circuit, listed by the duty generator, is *E* = *E*p sin ω*t*, and the current is *i* = *i*p sin (ω*t* + φ), whereby the subscript “p” means “peak,” and also φ is a phase factor. Because all the components are in series, the complete potential throughout the circuit equates to the sum of the potentials across the separation, personal, instance components, or *E* = *V**R* + *V**C* + *V**L*. Currently we must find the potential difference throughout each component, and the present in the circuit.

Perhaps the simplest means to perform this is to take it each component in turn, dealing with it together if it were the only component in the circuit. Beginning with the resistor, we have actually *VR* = *E*p sin ω*t*, which also equals *iRR*. Combining these gives us *iR* = (*E*p/*R*) sin ω*t*. This tells us that the voltage and current in the resistor vary in phase v each other.

For the capacitor, we have *VC* = *E*p sin ω*t*, and by the an interpretation for capacitance, *VC* = *q*/*C*. Together, these provide *q* = *E*p*C* sin ω*t*. Since current, *iC* = *dq*/*dt*, the present in the capacitor is *iC* = ω*CE*p cos ω*t*. This shows that the voltage and also current room out the phase v each other by 90°, or one quarter cycle, and that the voltage, i m sorry starts in ~ zero when the existing is in ~ its maximum, lags the current. If we set φ = -90° in the equation for existing in the very first paragraph above, and also then usage the enhancement formula for sines, we get *i* = *i*p cos ω*t*. To make this similar to the equation because that the resistor, we can write it together *iC* = (*E*p/*XC*) cos ω*t*, wherein *XC* = 1/ω*C*. *XC* is the *capacitive reactance*. It is analogous to resistance, and also its units are ohms. We deserve to see from the equations over that the preferably voltage throughout the capacitor is *VC*, p = *i*p*XC*.

For the inductor, we have *VL* = *E*p sin ω*t*, and also by the meaning for inductance, *VL* = *L*(*di*/*dt*). Together, these provide *di* = (*E*p/*L*) sin ω*t dt*. Therefore the current, *iL*, is the integral the this, or *iL* = -(*E*p/ω*L*) cos ω*t*. The voltage and also current space again out of phase by 90°, or one 4 minutes 1 cycle. This time, as soon as the voltage starts at zero, the present is an unfavorable and reaches its preferably one quarter cycle after the voltage has reached its maximum. For this reason the voltage leader the current, and the phase factor is +90°. Putting this right into the very first equation gives us *i* = -*i*p cos ω*t*. As we did for the capacitor, us can actors this equation as *iL* = -(*E*p/*XL*) cos ω*t*, wherein *XL* = ω*L*. *XL* is the *inductive reactance*. That is likewise analogous come resistance, and its systems are likewise ohms. We also see from the equations the the maximum voltage throughout the inductor is *VL*, p = *i*p*XL*.

Now us must find the existing in the entirety circuit. We know that the voltages must include according come *E* = *V**R* + *V**C* + *V**L*. Since these voltages every vary v time and have different phases v respect to each other, we cannot simply include the best voltages for which we derived expressions above, then divide to acquire the current. One means to address this is to stand for each voltage/current pair (*i.e.*, *V* and *i* for each component) as a pair the *phasors*, the is, vectors that revolve counterclockwise v angular frequency ω, who magnitudes space the height values that the voltage and also current, and whose projections on the *y*-axis offer the instantaneous values of voltage and also current. The algebraic amount of these instantaneous projections provides *E*. Your vector sum offers *E*p. *VC* and *VL* room 180° apart, and also both room 90° come *V**R*. Therefore, we can obtain *E*p by *E*p = √(*VR*2 + (*V**L* - *V**C*)2), whereby all the voltages space the optimal voltages. This amounts to √((*i*p*R*)2 + (*i*p*XL* - *i*p*XC*)2), and we obtain *E*p = *i*p√(*R*2 + (*XL* - *XC*)2). The quantity that contains the resistance, inductive reactance and also capacitive reactance is the *impedance*, typically denoted *Z*, and also *i*p = *E*p/*Z*. If we substitute the expressions because that the inductive and also capacitive reactances and rearrange, we attain *i*p = *E*p/√(*R*2 + (ω*L* - 1/ω*C*)2).

From the relationships among the phasors, we acquire the phase angle, φ, for the current in the circuit, indigenous tan φ = (*VL* - *VC*)/*VR*, where, again, every voltages are peak voltages. This equals *i*(*XL* - *XC*)/*iR* = (*XL* - *XC*)/*R*.

Now we an alert something interesting about the expression because that the present in the LRC circuit, which is the the existing has a maximum as soon as the inductive reactance equals the capacitive reactance, or ω*L* = 1/ω*C*. At this allude the inductive and also capacitive state in the expression because that *Z* cancel, and the impedance becomes completely resistive. This condition, where the current is a maximum, is called *resonance*. By rearranging the preceding equality, we uncover that the obtains once ω = 1/√*LC*. For our circuit, *L* = 3 mH, *C* = 220 pF, and 1/√*LC* = 1.2 × 106 rad/s, or, splitting by 2π, 200 kHz. At this point, the expression over for maximum current becomes *i*p = *E*p/*R*, or, generally, *i* = *E*p sin ω*t*/*R*. Keep in mind that at resonance, since *XL* = *XC*, the phase angle because that the present is zero.

Now, at long last, we acquire to the oscilloscope traces that highlight the operations of this circuit. Listed below are two oscilloscope displays taken at resonance. The peak trace (blue) is the voltage used by the role generator. The middle trace (pink) is *VR*. The bottom map (light purple) is either *VC* (left) or *VL* (right). The frequency at resonance is 195 kHz, nearby to that calculated over from the nominal ingredient values.

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For all of the traces shown below, the change resistor was set to 1 kΩ.