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Classwork Series and Exercises { Physics SS3}: AC IN RESISTOR, INDUCTOR AND CAPACITOR

Physics SS 3 Week 2

Topic: AC IN RESISTOR, INDUCTOR AND CAPACITOR

Alternating Current

Direct current (DC) circuits involve current flowing in one direction. In alternating current (AC) circuits, instead of a constant voltage supplied by a battery, the voltage oscillates in a sine wave pattern, varying with time as V = Vo sin ωt.

In a household circuit, the frequency is 60 Hz. The angular frequency is related to the frequency, f, by ω = 2πfVo represents the maximum voltage, which in a household circuit in North America is about 170 volts. We talk of a household voltage of 120 volts, though; this number is a kind of average value of the voltage. The particular averaging method used is something called root mean square (square the voltage to make everything positive, find the average, take the square root), or rms. Voltages and currents for AC circuits are generally expressed as r.m.s. values. For a sine wave, the relationship between the peak and the r.m.s. average is:

r.m.s. value = 0.707 peak value

resistance

Resistance in an AC circuit

The relationship V = IR applies for resistors in an AC circuit, so I = V/R = (Vo/R) sin(ωt) = Io sin(ωt)

In AC circuits we’ll talk a lot about the phase of the current relative to the voltage. In a circuit which only involves resistors, the current and voltage are in phase with each other, which means that the peak voltage is reached at the same instant as peak current. In circuits which have capacitors and inductors (coils) the phase relationships will be quite different.

Capacitance in an AC circuit

capacitance

Consider now a circuit which has only a capacitor and an AC power source (such as a wall outlet). A capacitor is a device for storing charging. It turns out that there is a 90° phase difference between the current and voltage, with the current reaching its peak 90° (1/4 cycle) before the voltage reaches its peak. Put another way, the current leads the voltage by 90° in a purely capacitive circuit.

To understand why this is, we should review some of the relevant equations, including: relationship between voltage and charge for a capacitor:

CV = Q relationship between current and the flow of change: I = ∆Q/∆t

The AC power supply produces an oscillating voltage. We should follow the circuit through one cycle of the voltage to figure out what happens to the current.

Step 1 – At point a (see diagram) the voltage is zero and the capacitor is uncharged. Initially, the voltage increases quickly. The voltage across the capacitor matches the power supply voltage, so the current is large to build up charge on the capacitor plates. The closer the voltage gets to its peak, the slower it changes, meaning less current has to flow. When the voltage reaches a peak at point b, the capacitor is fully charged and the current is momentarily zero.

Step 2 – After reaching a peak, the voltage starts dropping. The capacitor must discharge now, so the current reverses direction. When the voltage passes through zero at point c, it’s changing quite rapidly; to match this voltage the current must be large and negative.

Step 3 – Between points c and d, the voltage is negative. Charge builds up again on the capacitor plates, but the polarity is opposite to what it was in step one. Again the current is negative, and as the voltage reaches its negative peak at point d the current drops to zero.

Step 4 – After point d, the voltage heads toward zero and the capacitor must discharge. When the voltage reaches zero it’s gone through a full cycle so it’s back to point a again to repeat the cycle.

The larger the capacitance of the capacitor, the more charge has to flow to build up a particular voltage on the plates, and the higher the current will be. The higher the frequency of the voltage, the shorter the time available to change the voltage, so the larger the current has to be. The current, then, increases as the capacitance increases and as the frequency increases.

Usually this is thought of in terms of the effective resistance of the capacitor, which is known as the capacitive reactance, measured in ohms. There is an inverse relationship between current and resistance, so the capacitive reactance is inversely proportional to the capacitance and the frequency:

A capacitor in an AC circuit exhibits a kind of resistance called capacitive reactance, measured in ohms. This depends on the frequency of the AC voltage, and is given by Capacitive reactance Xc = 1/ωC = 1/2πfC

We can use this like a resistance (because, really, it is a resistance) in an equation of the form V = IR to get the voltage across the capacitor:

V = IXc

Note that V and I are generally the r.m.s. values of the voltage and current.

Inductance in an AC circuit

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An inductor is simply a coil of wire (often wrapped around a piece of ferromagnet). If we now look at a circuit composed only of an inductor and an AC power source, we will again find that there is a 90° phase difference between the voltage and the current in the inductor. This time, however, the current lags the voltage by 90°, so it reaches its peak 1/4 cycle after the voltage peaks.

The reason for this has to do with the law of induction:

e = -N∆ɸ/∆t or e = – L∆I/∆t

Applying Kirchoff’s loop rule to the circuit above gives:

V – L∆I/∆t = 0 so V = L∆I/∆t

As the voltage from the power source increases from zero, the voltage on the inductor matches it. With the capacitor, the voltage came from the charge stored on the capacitor plates (or, equivalently, from the electric field between the plates). With the inductor, the voltage comes from changing the flux through the coil, or, equivalently, changing the current through the coil, which changes the magnetic field in the coil.

To produce a large positive voltage, a large increase in current is required. When the voltage passes through zero, the current should stop changing just for an instant. When the voltage is large and negative, the current should be decreasing quickly. These conditions can all be satisfied by having the current vary like a negative cosine wave, when the voltage follows a sine wave.

How does the current through the inductor depend on the frequency and the inductance? If the frequency is raised, there is less time to change the voltage. If the time interval is reduced, the change in current is also reduced, so the current is lower. The current is also reduced if the inductance is increased.

As with the capacitor, this is usually put in terms of the effective resistance of the inductor. This effective resistance is known as the inductive reactance. This is given by XL = ωL = 2πfL, where L is the inductance of the coil (this depends on the geometry of the coil and whether it’s got a ferromagnetic core). The unit of inductance is the henry.

As with capacitive reactance, the voltage across the inductor is given by: V = IXL

Where does the energy go?

One of the main differences between resistors, capacitors, and inductors in AC circuits is in what happens with the electrical energy. With resistors, power is simply dissipated as heat. In a capacitor, no energy is lost because the capacitor alternately stores charge and then gives it back again. In this case, energy is stored in the electric field between the capacitor plates. The amount of energy stored in a capacitor is given by energy in a capacitor: Energy = ½ CV2

In other words, there is energy associated with an electric field. In general, the energy density (energy per unit volume) in an electric field with no dielectric is: Energy density in an electric field = ½ e0 E2

With a dielectric, the energy density is multiplied by the dielectric constant.

There is also no energy lost in an inductor, because energy is alternately stored in the magnetic field and then given back to the circuit. The energy stored in an inductor is:

Energy in an inductor: Energy = ½ LI2

Again, there is energy associated with the magnetic field. The energy density in a magnetic field is Energy density in a magnetic field = B2/(2μ0).

Power in an A.C. Circuit

The average power in an a.c. circuit is given by

P = IVcosɸ

where I, V, are the effective (r.m.s.) values of the current and voltage respectively and ɸ is the angle of lag or lead between them. The quantity cos ɸ is known as the power factor of the device. The power factor can have any value between zero and unity for ɸ varying from 90o (or cos ɸ = 0) average power P, is zero. A power of zero means that the device is a pure reactance, inductance or capacitance. Thus no power is dissipated in an inductance or capacitance. However if I is the r.m.s. value of the current in a circuit containing a resistance, R, then the power absorbed in the resistance is given by

P = I2R

For an a.c. circuit, the instantaneous power is given by P =IV (instantaneous values.)

Cos ɸ = R/Z = Resistance/Impedance

Resonance in RLC Series Circuit

The current, I, in an RLC series circuit is given by

I = V/Z = V/√R2 + (XL – XC)2

Where XL = 2πfL and XC = 1/2πfC. The maximum current is obtained in the circuit when the impedance Z, in the above equation is minimum. This happens when

XL = XC or 2πfL = 1/2πfC

Resonance is said to occur in a.c. series circuit when the maximum current is obtained from such a circuit.

The frequency at which this resonance occurs is called the resonance frequency (f0). This is the frequency at which XL = XC or 2πf0L = 1/2πf0C.

Hence solving the above equation we obtain that f0 is given by

f0 = 1/2π√LC

or since ω = 2πf, we can write the condition of resonance as:

ω0 = 1/√LC

The variation of current I, and frequency f, in an RLC series circuit

Application of Resonance

The resonant circuit finds applications in electronics. It is used to tune radios and TVs. Its great advantage is that it responds strongly to one particular frequency. The other frequencies are very little effect.

Hence such a resonant circuit can select one signal of a definite frequency from a jumble of other signals available to it. That resonance frequency corresponds to that of a particular incoming radio signal. When this happens, maximum current is obtained and the distant radio station is loudly and clearly heard.

Example

An a.c. voltage of amplitude 2.0 volts is connected to an RLC series circuit. If the resistance in the circuit is 5 ohms, and the inductance and capacitance are 3 mH and 0.05μF respectively, calculate

i. the resonance frequency f0

ii. the maximum a.c. current at resonance

Solution

f0 = 1/2π√LC

   = 1/2π√3 x 10-3 x 5 x 10-8 = 1/2π√15 x 10-11

   = 1299.545 Hz

   = 13 KHz

At resonance Z = R since XL = XC

I0 = V0/R = 2/5 = 0.4 Amps

CAPACITIVE REACTANCE

So far you have been dealing with the capacitor as a device which passes ac and in which the only opposition to the alternating current has been the normal circuit resistance present in any conductor. However, capacitors themselves offer a very real opposition to current flow. This opposition arises from the fact that, at a given voltage and frequency, the number of electrons which go back and forth from plate to plate is limited by the storage ability-that is, the capacitance-of the capacitor. As the capacitance is increased, a greater number of electrons change plates every cycle, and (since current is a measure of the number of electrons passing a given point in a given time) the current is increased.

 Increasing the frequency will also decrease the opposition offered by a capacitor. This occurs because the number of electrons which the capacitor is capable of handling at a given voltage will change plates more often. As a result, more electrons will pass a given point in a given time (greater current flow). The opposition which a capacitor offers to ac is therefore inversely proportional to frequency and to capacitance. This opposition is called CAPACITIVE REACTANCE.

You may say that capacitive reactance decreases with increasing frequency or, for a given frequency, the capacitive reactance decreases with increasing capacitance. The symbol for capacitive reactance is XC.

Now you can understand why it is said that the XC varies inversely with the product of the frequency and capacitance. The formula is:

Xc = 1/2πfc

Where: XC is capacitive reactance in ohms f is frequency in Hertz C is capacitance in farads 2π is 6.28 (2 X 3.1416).

The following example problem illustrates the computation of XC.

Given : f = 100 Hz, C = 50μF

Solution: Xc = 1/2πfc

                   = 1/6.28 x 100Hz x 50μF

                         = 1/0.0314Ω

                   = 31.8Ω or 32Ω

REACTANCE, IMPEDANCE, AND POWER RELATIONSHIPS IN AC CIRCUITS

Up to this point inductance and capacitance have been explained individually in ac circuits. The rest of this part will concern the combination of inductance, capacitance, and resistance in ac circuits.

To explain the various properties that exist within ac circuits, the series RLC circuit will be used. Figure (4) is the schematic diagram of the series RLC circuit. The symbol shown in figure 4-4 that is marked E is the general symbol used to indicate an ac voltage source.

series rlc circuit

REACTANCE

The effect of inductive reactance is to cause the current to lag the voltage, while that of capacitive reactance is to cause the current to lead the voltage. Therefore, since inductive reactance and capacitive reactance are exactly opposite in their effects, what will be the result when the two are combined? It is not hard to see that the net effect is a tendency to cancel each other, with the combined effect then equal to the difference between their values. This resultant is called REACTANCE; it is represented by the symbol X; and expressed by the equation X = XL – XC or X = XC – XL. Thus, if a circuit contains 50 ohms of inductive reactance and 25 ohms of capacitive reactance in series, the net reactance, or X, is 50 ohms – 25 ohms, or 25 ohms of inductive reactance.

For a practical example, suppose you have a circuit containing an inductor of 100 µH in series with a capacitor of .001 µF, and operating at a frequency of 4 MHz What is the value of net reactance, or X?

Given f = 4MHz, L = 100μH, C =0.001μF

Solution: XL = 2πfL

                     = 6.28 x 4MHz x 100μH = 2512 Ω

                Xc = 1/2πfC

                     = 1/6.28 x 4MHz x 0.001μF

                     = 1/0.2515Ω

                     = 39.8Ω

                 X = XL – XC

                 X = 24722Ω (inductive)

IMPEDANCE

From your study of inductance and capacitance you know how inductive reactance and capacitive reactance act to oppose the flow of current in an ac circuit. However, there is another factor, the resistance, which also opposes the flow of the current. Since in practice ac circuits containing reactance also contain resistance, the two combine to oppose the flow of current. This combined opposition by the resistance and the reactance is called the IMPEDANCE, and is represented by the symbol Z.

Since the values of resistance and reactance are both given in ohms, it might at first seem possible to determine the value of the impedance by simply adding them together. It cannot be done so easily, however. You know that in an ac circuit which contains only resistance, the current and the voltage will be in step (that is, in phase), and will reach their maximum values at the same instant. You also know that in an ac circuit containing only reactance the current will either lead or lag the voltage by one-quarter of a cycle or 90 degrees. Therefore, the voltage in a purely reactive circuit will differ in phase by 90 degrees from that in a purely resistive circuit and for this reason reactance and resistance are combined by simply adding them.

When reactance and resistance are combined, the value of the impedance will be greater than either. It is also true that the current will not be in step with the voltage nor will it differ in phase by exactly 90 degrees from the voltage, but it will be somewhere between the in-step and the 90-degree out-of-step conditions. The larger the reactance compared with the resistance, the more nearly the phase difference will approach 90°. The larger the resistance compared to the reactance, the more nearly the phase difference will approach zero degrees.

If the value of resistance and reactance cannot simply be added together to find the impedance, or Z, how is it determined? Because the current through a resistor is in step with the voltage across it and the current in a reactance differs by 90 degrees from the voltage across it, the two are at right angles to each other. They can therefore be combined by means of the same method used in the construction of a right-angle triangle.

Assume you want to find the impedance of a series combination of 8 ohms resistance and 5 ohms inductive reactance. Start by drawing a horizontal line, R, representing 8 ohms resistance, as the base of the triangle.

Then, since the effect of the reactance is always at right angles, or 90 degrees, to that of the resistance, draw the line XL, representing 5 ohms inductive reactance, as the altitude of the triangle. This is shown in figure (5). Now, complete the hypotenuse (longest side) of the triangle. Then, the hypotenuse represents the impedance of the circuit.

Vector diagram showing relationship of resistance, inductive reactance, and impedance in a series circuit.

vector

One of the properties of a right triangle is:

(hypotenuse)2 = (base)2 + (altitude)2

Hypotenuse = √(base)2 (altitude)2

Applied to impedance, this becomes,

(impedance)2 = (resistance)2 + (reactance)2

Impedance = √(resistance)2 + (reactance)2

Impedance = √(resistance) 2 + (reactance)2

Z = √R2 + X2

Now suppose you apply this equation to check your results in the example given above.

Given : R = 8Ω, XL =5Ω

Solution : Z = √R2 + XL 2

                   = √(8Ω)2 + (5Ω)2

                                 = √64 + 25

                     = √89Ω = 9.4Ω

When you have a capacitive reactance to deal with instead of inductive reactance as in the previous example, it is customary to draw the line representing the capacitive reactance in a downward direction. This is shown in figure (6). The line is drawn downward for capacitive reactance to indicate that it acts in a direction opposite to inductive reactance which is drawn upward. In a series circuit containing capacitive reactance the equation for finding the impedance becomes:

Z = √R2 + XC2

vector 2

In many series circuits you will find resistance combined with both inductive reactance and capacitive reactance. Since you know that the value of the reactance, X, is equal to the difference between the values of the inductive reactance, XL, and the capacitive reactance, XC, the equation for the impedance in a series circuit containing R,XL, and XCthen becomes:

Z = √R2 + (XL – XC)2 or Z = √R2 + X2

Note: The formulas Z = √R2 + XC2, Z = √R2 + X2 and Z = √R2 + XL 2 can be used to calculate Z only if the reisitance and reactance are connected in series.

In the above expressions, you will see the method which may be used to determine the impedance in a series circuit consisting of resistance, inductance, and capacitance.

vector3

Assume that 10 ohms inductive reactance and 20 ohms capacitive reactance are connected in series with 40 ohms resistance. Let the horizontal line represent the resistance R. The line drawn upward from the end of R, represents the inductive reactance, XL. Represent the capacitive reactance by a line drawn downward at right angles from the same end of R. The resultant of XL and XC is found by subtracting XLfrom XC. This resultant represents the value of X.

Thus: X = XC – XL = 20 ohms – 10 ohms

               = 10 ohms

Note: If inductive reactance is smaller than the capacitive reactance and is therefore subtracted from the capacitive reactance.

These two examples serve to illustrate an important point: when capacitive and inductive reactance are combined in series, the smaller is always subtracted from the larger and the resultant reactance always takes the characteristics of the larger.

Question

1. Given XL = 10 Ω, XC = 20 Ω, R = 40 Ω, calculate the impedance.

A. 40Ω B. 41.2Ω C. 56Ω D. 45.3Ω

Given f = 1MHz, L = 100μH, C = 0.0002μF

2. Calculate for inductive reactance

A. 352Ω B. 702Ω C. 628Ω D. 245Ω

3. Calculate for capacitive reactance

A. 801Ω B. 235Ω C. 769Ω D. 745Ω

4. Now assume you have a circuit containing a 100 – µH inductor in series with a .0002-µF capacitor, and operating at a frequency of 1 MHz. What is the value of the resultant reactance in this case?

A. 168Ω B. 200Ω C. 265Ω D. 158Ω

5. As with capacitive reactance, the voltage across the inductor is given by

A. IXL  B. IXC C. IXR D. IXZ

Answer

1. B 2. C 3. C 4. A 5. A

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1 thought on “Classwork Series and Exercises { Physics SS3}: AC IN RESISTOR, INDUCTOR AND CAPACITOR”

  1. This is awesome. Great work. I love this, as a Physics teacher, I’ve found another teacher in this site. Thank you so much. So proud of you.

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