Chapter 33

AC Circuits:

In this chapter we briefly study the behavior of resistors, capacitors and inductors in series with an ac source.  First, each of these elements alone with an ac source will be studied and then a combination of them in series with an ac source will be considered.

A Resistor and an AC Source:

A resistor connected to an ac source acts similar to the case of being connected to a dc source.  The symbol for an ac source is as shown in the figure The voltage-current relation (Ohm's law) is:

 Vrms = R Irms    and    Vmax = R Imax.

 

The voltage across and the current 

through a resistor are in phase. They reach maximum, minimum, and zero together.

 

This figure shows an instant at which the top of the source is positive and the bottom is negative.

Example 1:  In the above figure, suppose V = 17.0sin(377t) and R = 34.0Ω.  Find the rms current.

Solution:  From the given equation:  Vmax = 17.0 volts   &  ω = 377 rd/s.

Imax = Vmax /R;    Imax = 0.500A ;     Irms = 0.707Imax ;     Irms = 0.354A.

 

A Capacitor and an AC Source: 

When a capacitor is connected to an ac source, in half a cycle it faces charging and in the next half-cycle faces discharging.  This develops a certain resistance in the circuit that is called "capacitive reactance, Xc."  Capacitive reactance depends on two factors.  One is the capacity C itself.  The greater the capacity C, the less reactive the capacitor is toward being charged or discharged.  The other factor is the angular frequency w of the source that alternates charging and discharging.  The less the frequency of the source, the more time the capacitor has in each cycle to charge or discharge and therefore develops more resistance.  The formula for capacitive reactance, Xc and a circuit diagram for a capacitor connected to an ac source are shown below.  The unit of Xc in SI is Ω.   The formula for the capacitor current is also shown below.  Note that contrary to a resistor, the current through a capacitor is not in phase with the voltage across it. 

 

The current leads voltage by 90o when a capacitor is connected to an ac source.

 

 

 

Comparing the voltage and current graphs, note that when VC is zero, IC is maximum and when IC is zero, Vis maximum as expected.  That is why there is a phase difference of 90o between the two.

Example 2:  In the above figure, suppose V = 8.50sin(314t) and C = 63.7μF.  Find the rms current.

Solution:  From the equation:  Vmax = 8.50volts  and  ω = 314 rd/s.

XC = 1/(Cω)    ;    XC = 1 / [(63.7x10-6F)(314rd/s)]  =  50.0 Ω

Imax = Vmax /XC ;    Imax = 0.170A ;    Irms = 0.707Imax ;    Irms = 0.120A.

 

An Inductor and an AC Source:

When an inductor is connected to an ac source, the inductor reacts to the nonstop changes in the current and develops a nonstop opposing voltage to resist the changes.  This opposing voltage creates a type of resistance in the circuit that is called "inductive reactance, XL."   Inductive reactance depends on two factors: one is the self-inductance L of the inductor.  The greater the L of the inductor, the greater resistance it develops toward the changes in the current.  The other factor is the angular frequency w of the source.  The greater the frequency of the source, the faster the changes are made in the current and therefore the greater the inductor reactance.  The formula for inductive reactance and a circuit diagram for an inductor connected to an ac source are shown below.  The SI unit of XL is Ω.   The formula for the current through the inductor is also shown below.  Note that contrary to a resistor, the current through an inductor is not in phase with the voltage applied across it. 

The current lags voltage by 90o when an inductor is connected to an ac source.

 

 

This figure shows an instant at which the top of the source is positive and the bottom is negative.

 

Comparing the voltage and current graphs, note that when VL is zero, IL is maximum and when IL is zero, VL  is maximum as expected.  That is why there is a phase difference of 90o between the two.

 

Example 3:  In the above figure, suppose V = 36sin(720t) and L = 250mH.  Find the rms current.

Solution:  From the equation: Vmax = 36 volts  and  ω =720 rd/s.

XL = Lω    ;    XL = (0.25H)(720 rd/s)  =  180 Ω

Imax = Vmax/XL ;    Imax = 0.20A.     ; Irms = 0.707Imax ;    Irms = 0.14 A.

 

The General Case: The RCL Series Circuit:

  The general case is when an ac source is in series with a resistor, a capacitor, and an inductor.  The sequence is not important.  In this case, it can be shown mathematically that the overall resistance called the "impedance, Z" of the circuit is given by the formula shown on the right.  Again, the unit of Z turns out to be Ω.  The amount that current may lead or lag voltage depends on capacity C, inductance L, and resistance R of the elements.  Mathematical solution provides an angle φ that measures the amount of lead or lag between the current and voltage in the circuit.  A positive φ indicates that the voltage is leading the current by φ.  The formula for φ is also given on the right.  The max and rms currents may be found from

Imax = Vmax/Z   &   Irms = Vrms/Z.

This figure shows an instant at which the top of the source is positive and the bottom is negative.

Note:  φ is pronounced " phi."

 

Example 4: In the above figure, let L = 25.0mH, C = 125μF, R = 28.3Ω, and V = 18.0sin(400t).  Find the Irms and the phase angle φ.

Solution:  From the equation:  Vmax = 18.0 volts  and  ω = 400rd/s.

XL = Lω    ;    XL = (0.025H)(400 rd/s)  = 10.0 Ω

XC = 1/(Cω) ;    XC = 1/[(125x10-6)(400)] = 20.0Ω

Z = [R2 + (XL - XC)2 ]1/2  =  [28.32 + (10 - 20)2 ]1/2  =  30.0Ω

Imax = Vmax/Z ;    Imax = 0.600A ;     Irms = 0.707Imax ;    Irms = 0.424A.

φ = tan-1 [ (XL- XC )/R ]  =  tan-1[-0.353]  = -19.4o.

LC Resonance Circuits:

If R = 0 in an RCL circuit, the circuit forms the so-called "LC circuit."   LC circuits are used as electric oscillators.   Each LC-circuit has a natural frequency that depends on the values of capacitance C and inductance L.  The following figure shows a LC circuit in which the capacitor has an initial charge Qo on it.  As soon as key K is turned on and the circuit is closed, charge Qo tends to flow into the inductor creating a varying current in it.  The inductor reacts strongly toward the change in the current, and returns the current back to the capacitor.  The capacitor getting charged again returns the current toward the inductor and the process keeps repeating.  If there is no ohmic resistance to dissipate energy, theoretically this should continue for ever and the result is an electric oscillator.  This is similar to a mass-spring system (with no friction) that is put into oscillation.  It oscillates for a long time.

 

Example 5:  In the above figure, suppose C = 250nF and initially charged, and L = 4.0μH.  Find (a) the natural angular frequency ω  (b) the natural frequency f, and (c) the period T of the oscillations.

Solution: (a) ω = 1/(LC)1/2 1/[(250x10-9F)(4.0x10-6H)] = 1.0x106 rd/s.

(b) ω = 2π f ;    f = w/2π ;    f  = 1.0x106/(2π) = 160,000Hz  or 160kHz.

(c) T = 1/;    T = 1/160,000s-1  =  6.28x10-6 s.

An Application of LC Circuit:

The LC combination is very important in transmitter or receiver circuits.  The figure on the right shows an LC circuit that is coupled at L1L2 with another inductor that receives signals from an antenna.  Inductor L1 in the antenna circuit induces a current in inductor L2 of the LC circuit.  The capacity C in the LC circuit can be changed such that the natural frequency of the LC circuit becomes equal to the frequency of the received signal.  When this tuning occurs, the LC circuit is said to be in resonance with the received signal frequency.  The antenna receives a very large number of signals at a wide span of frequencies coming in from all directions.   The LC circuit picks up the one signal that matches its natural frequency.  The LC circuit can be adjusted to a desired natural frequency; in other words,  the receiver (the radio set) can be tuned to a desired station with this simple circuit. 

The tuning knob on some radio sets changes the capacity of the variable capacitor shown in the figure.

 

Example 6:  The frequency used for communications in cell phones is about 2000MHz or 2.0GHz.  Find (a) the angular frequency w for such signals.  (b) If in the receiver a 0.050pF-capacitor is used, find the inductance L of the inductor to be used to create resonance at that frequency.

Solution: (a) ω = 2πf ;   ω = 2π(2.0x109Hz) = 1.3x1010rd/s.

(b) ω = 1/(LC)1/2 ;     ω2 =  1/LC ;    L = 1/[Cw2]  = 1.27x10-7H

 

Example 7: In the previous example, suppose we want to make that inductor (coil).  How many loops of a thin copper wire should be wrapped around a thin cylinder 5.0mm in diameter if the length of the resulting coil is not to exceed 2.0cm.    μo = 4π x 10-7 Tm /A.

Solution:  The formula for self-inductance L is:   L = μon2 A l   in which n is the number of turns per meter.  Solving for n2 yields:   n2  =  L/(μoA l).

n2  = 1.27x10-7 /[(4π x 10-7 Tm/A)(π)(.0025m)2 (.020m)] = 257355 (turn/m)2

n = 507turns/;    N = nl ;    N = 507(turns / m)( .020m) = 10 turns.

 

Test Yourself 1:   click here.

1)  A resistor connected to an AC source (a) behaves similar to being connected to a DC source   (b) the voltage-current relation or Ohm's law is : Vrms = R Irms  (c) the voltage-current relation or Ohm's law is : Vmax = R Imax   (d) a,b, & c.

2) For a  resistor R connected to an AC source (a) I and V are in phase  (b) I  leads  V by 90o   (c) I  lags V by 90o.

3) A light bulb with a hot resistance of 144Ω connected to a wall outlet, pulls a rms current of (a) 1.20A   (b) 0.83A   (c) 0.95A.   click here.

4) A light bulb with a hot resistance of 195Ω connected to a wall outlet, pulls a maximum current of (a) 1.20A    (b) 0.83A    (c) 0.87A.

5) Across a resistor connected to an AC source, when the voltage is maximum , the current is (a) minimum  (b)  zero   (c)  maximum.

6) The capacitor current when connected to an AC source, (a) lags the voltage by 90o   (b) leads the voltage by 90o   (c) leads the voltage by 45o.   click here.

7) The capacitive reactance of a capacitor depends on (a) capacity C only  (b) the angular frequency ω of the AC source only   (c) both C and ω.

8) The formula that calculates the capacitive reactance XC is (a) XC = 1/(Cω)   (b)  XC = Cω    (c) neither a nor b.

9) The unit of XC is (a) Ohm   (b) Ohm/s   (c) Ohm-sec.   click here.

Problem: A 36-μF capacitor is connected to an AC source with  equation V = 170sin(100π t) where V is in volts and t in seconds.   Draw a diagram for the problem and answer the following questions (10 through 14):

10) The maximum voltage is (a) 120V   (b) 170V   (c) 83V.   click here.

11)  The angular frequency is (a) 100π rd/s     (b) 314 rd/s    (c) a & b.

12) The capacitive reactance  XC  is (a) 44Ω   (b) 66Ω    (c) 88Ω.

13) The maximum current is (a) 19A    (b) 1.9A    (c) 0.85A.

14) The rms current is (a) 1.4A    (b) 2.7A    (c) 3.7A.   click here.

15) For an inductor in an AC source, current (a) leads voltage by 90o   (b) leads voltage by 45o   (c) lags voltage by 90o.

16) The inductive reactance  XL  of an inductor depends on (a) just the inductance L of the inductor only   (b) the angular frequency  ω  of the AC source it is connected to only  (c) both L & ω.   click here.

17) The formula that calculates the inductive reactance XL is (a) XL = 1/(Lω)    (b)  XL = Lω    (c) neither a nor b.

18) The unit of XL is (a) Ohm/s   (b) Ohm-sec   (c) Ohm.

Problem: A 0.16-H inductor is connected to an AC source of equation V = 68sin(120π t) where V is in volts and t in seconds.  Draw a diagram for the problem and answer the following questions:

19) The maximum voltage is (a) 68V   (b) 134V   (c) 96V.   click here.

20)  The angular frequency is (a) 100π rd/s     (b) 377 rd/s    (c) a & b.

21) The inductive reactance  XL  is (a) 30.Ω   (b) 60.Ω    (c) 120Ω.

22) The maximum current is (a) 1.1A    (b) 1.6A    (c) 2.8A.   click here.

23) The rms current is (a) 1.4A    (b) 2.7A    (c) 0.80A.

Problem: A 250mH inductor, a 63μF capacitor, and a 94Ω resistor are series with an AC source for which voltage variations is given by V = 34sin(48πt).  Draw a diagram for the problem and answer the following questions:

24) The maximum voltage is (a) 68V   (b) 34V   (c) 96V.     click here.

25) The angular frequency is (a) 48π rd/s     (b) 151 rd/s    (c) a & b.

26) The capacitive reactance  XC  is (a) 330Ω   (b) 60Ω    (c) 105Ω.

27) The inductive reactance  XL  is (a) 38Ω   (b) 76 Ω    (c) 120Ω.   click here.

28) The impedance  Z  of the circuit is (a) 11.3Ω   (b) 76.Ω    (c) 115Ω.

29) The maximum current in the circuit is (a) 0.30A    (b) 1.3A    (c) 2.6A.

30) The rms current is (a) 0.21A    (b) 1.1A    (c) 2.2A     click here.

31) An electric oscillator (resonator) is (a) an RCL circuit with no resistance (R = 0) in it   (b) is an LC circuit in which the ohmic resistance of the inductor is negligible   (c) both a & b.

32) If R is zero in the formula for impedance Z, we may write: (a) XC - XL = 0    (b) Lω = 1/(Cω)    (c) both a & b.

33) In Question 32, solving for ω by cross-multiplication, yields: (a) LCω2 = 1    (b) ω2 = 1/ (LC)     (c) a & b.   click here.

Problem:  Suppose you want to make an LC resonator with a natural frequency of 1850MHz.   If the capacity of the available capacitor is 0.060pF, follow the steps asked below to find the inductance L needed to make the LC resonator.

34) w is (a) 1.16x1010rd/s      (b) 5.8x109 rd/s     (c) 1.62x1010rd/s.

35) Solving for L from ω2 = 1/(LC) yields: (a) 1.238x10-7H   (b) 2.81x10-6H    (c) 3.39x10-8H.   click here.

36) Use the equation L = μon2 A l  and suppose that the inductor's length and diameter are to be 1.8cm and 5.0mm respectively, and solving for n.   The number of turns per meter, n, is (a) 324   (a) 528    (a) 622.

37) The number of turns N is (a) 18    (b) 9.5    (c) 14.5 turns.