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:

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 maximum and rms currents.
Solution: From the equation of the source, it is clear that V_{max} = 17.0volts , and w = 377 rd/s.
I_{max} = V_{max} /R ; I_{max} = 0.500Amps ; I_{rms} = 0.707I_{max} ; I_{rms} = 0.354 Amps.
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 halfcycle faces discharging. This develops a
certain resistance in the circuit that is called "capacitive reactance, X_{c}."
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,
X_{c} and a circuit
diagram for a capacitor connected to an ac source are shown on the right.
The unit of X_{C} in SI is
Ω .
The formulas for the current through the circuit or the
capacitor is also shown on the right. Note that contrary to a resistor,
the current through a
capacitor is not in phase with voltage across
it.

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

In the voltage and current graphs, note that when the voltage across the capacitor is zero, the magnitude of the current is maximum as is expected and vice versa. That is why there is a phase difference of 90° between the two.
Example 2: In the above figure, suppose V = 8.50sin(314t) and C = 63.7μF. Find the maximum and rms currents.
Solution: From the equation of the source, it is clear that V_{max} = 8.50volts , and w = 314 rd/s.
X_{C} = 1 / (Cω) ; X_{C} = 1 / [(63.7x10^{6}F)(314rd/s)] = 50.0 Ω
I_{max} = V_{max} / X_{C} ; I_{max} = 0.170Amps ; I_{rms} = 0.707I_{max} ; I_{rms} = 0.120 Amps.
An Inductor and an AC Source:
When an inductor is connected to an ac source,
the inductor resists the changes in current due to Lenz's law and develops a
certain resistance in the circuit that is called "inductive reactance, X_{L}."
Inductive reactance depends on two factors. One is the inductance
L itself. The greater the
inductance L, the greater resistance it shows toward the changes in current. The other factor is the angular frequency
w of the source. The greater the frequency of the source, the
stronger it reacts and greater resistance it shows. The formula for
inductive reactance and a circuit
diagram for an inductor connected to an ac source are shown on the right.
The unit of X_{L} in SI is
Ω.
The formula for current through the circuit or the
inductor is also shown on the right. Note that
contrary to a resistor, the current through an
inductor is not in phase with the voltage applied across
it.

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

From the graphs it can be seen that the maximum change in the current occurs when it is passing through zero. It is at that instant that the developed opposite voltage reacts strongly and reaches its maximum. That is why there is a phase difference of 90° between the two.
Example 3: In the above figure, suppose V = 36sin(720t) and L = 250mH. Find the maximum and rms currents.
Solution: From the equation of the source, it is clear that V_{max} = 36 volts , and w = 720 rd/s.
X_{L} = Lω ; X_{L} = (0.25H)(720 rd/s) = 180 Ω
I_{max} = V_{max} / X_{L} ; I_{max} = 0.20Amps ; I_{rms} = 0.707I_{max} ; I_{rms} = 0.14 Amps.
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, Z has units of Ω. The amount that the current may lead or lag the voltage depends on the 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 I_{max} = V_{max} / Z and I_{rms} = V_{rms} / 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(400.t). Find the I_{rms} and the phase angle φ.
Solution: From the equation of the source, it is clear that V_{max} = 18.0 volts , and w = 400. rd/s.
X_{L} = Lω ; X_{L} = (0.025H)(400 rd/s) = 10.0 Ω
X_{C} = 1/Cω ; X_{C} = 1/[(125x10^{6})(400)] = 20.0Ω
Z = [ R^{2} + (X_{L}  X_{C})^{2} ]^{1/2} = [ 28.3^{2} + (10  20)^{2} ]^{1/2} = 30.0Ω
I_{max} = V_{max} / Z ; I_{max} = 0.600Amps ; I_{rms} = 0.707I_{max} ; I_{rms} = 0.424 Amps.
φ = tan^{1} [ ( X_{L}  X_{C } ) / R ] = tan^{1}[0.353] = 19.4º.
LC Resonance Circuits:
If R = 0 in a LCR circuit, the circuit forms the socalled " LC circuit. " LC circuits are used as electric oscillators. Each LCcircuit has a natural frequency that depends on the values of capacity C and inductance L. The following figure shows a LC circuit in which the capacitor has an initial charge Q_{o} on it. As soon as key K is turned on and the circuit is closed, charge Q_{o} 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 massspring system (with no friction) 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 w , (b) the natural frequency f, and (c) the period T of the oscillations.
Solution: (a) w = 1 / SQRT( LC ) = 1 / [(250x109F)(4.0x106H)] = 1.0x10^{6} rd/s
(b) w = 2π f ; f = w/2π ; f = 1.0x10^{6} / 2π = 160,000 Hz or 160 kHz
(c) T = 1 / f ; T = 1 / 160,000s^{1} = 6.28x10^{6} seconds.
An Application of LC Circuit:
The LC combination is very important in transmitter and receiver circuits. The figure on the right shows an LC circuit that is coupled (at L_{1}L_{2}) with another inductor that receives signals from an antenna. The inductor (L_{1}) in the antenna circuit induces a current in the inductor (L_{2}) 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. We can adjust the LC circuit to have the natural frequency we want; in other words, we can tune the receiver ( the radio set) to the desired station with this simple circuit. The tuning knob on most radio sets changes the capacity of that variable capacitor. 
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 0.050pFcapacitor is used, find the inductance L of the inductor to be used to create resonance at that frequency.
Solution: (a) w = 2π f ; w = 2π(2.0x10^{9} Hz) = 1.256x10^{10} rd/s or (1.3E10)rd/s
(b) w = 1 / SQRT( LC ) ; w^{2} = 1 / LC ; L = 1 / [Cw^{2}] = 1.27x10^{7} Henry
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. μ_{o} = 4π x 10^{7} Tm /A.
Solution: Recall the formula for the inductance of a coil: L = μ_{o}n^{2} A l in which n is the number of turns per meter. Solving for n^{2} yields: n^{2} = L / μ_{o}A l.
n^{2} = 1.27x10^{7} / [ (4π x 10^{7} Tm/A)(π)(.0025m)^{2} (.020m) ] = 257355 (turn / m)^{2}
n = 507 turns / m. ; 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 voltagecurrent relation or Ohm's law is : V_{rms} = R I_{rms}. (c) the voltagecurrent relation or Ohm's law is : V_{max} = R I_{max}. (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) min. (b) zero. (c) max.
6) The current through a capacitor 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) just the capacity, C of the capacitor. (b) the angular frequency, ω, of the AC source it is connected to. (c) both a & b.
8) The formula that calculates the capacitive reactance, X_{C}, is (a) X_{C} = 1/(Cω). (b) X_{C} = Cω. (c) neither a nor b.
9) The unit of X_{C} is (a) Ohm. (b) Ohm/s. (c) Ohmsec. click here.
Problem: A 36μF capacitor is connected to an AC source of 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) 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, X_{C}, 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 at an AC source, current (a) leads voltage by 90º. (b) leads voltage by 45º. (c) lags voltage by 90º.
16) The inductive reactance, X_{L}, of an inductor depends on (a) just the inductance, L, of the inductor. (b) the angular frequency, ω, of the AC source it is connected to. (c) both a & b. click here.
17) The formula that calculates the inductive reactance, X_{L}, is (a) X_{L} = 1/(Lω). (b) X_{L} = Lω. (c) neither a nor b.
18) The unit of X_{L} is (a) Ohm/s. (b) Ohmsec. (c) Ohm.
Problem: A 0.16H 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, X_{L}, 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 whose voltage variations is given by V = 34sin(48π t). 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, X_{C}, is (a) 330.Ω. (b) 60.Ω. (c) 105Ω.
27) The inductive reactance, X_{L}, 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) A LCcircuit of resonator, is (a) an RCL circuit with no resistance (R = 0) in it. (b) is a LC circuit in which the ohmic resistance of the inductor in it is negligible. (c) both a & b.
32) If R is zero in the formula for impedance, Z, we may write: (a) X_{C}  X_{L} = 0. (b) Lω = 1/(Cω). (c) a & b.
33) Solving for ω, by crossmultiplication, yields: (a) LCω^{2} = 1. (b) ω^{2} = 1 / LC. (c) a & b. click here.
Problem: We want to make a LC resonator whose natural frequency is 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) The angular frequency, w, is (a) 1.16x10^{10}rd/s. (b) 5.8x10^{9 }rd/s. (c) 1.62x10^{10}rd/s.
35) Solving for L from ω^{2} = 1 / LC yields: (a) 1.238x10^{7}H. (b) 2.81x10^{6}H. (c) 3..39x10^{8}H. click here.
36) Use the equation L = μ_{o}n^{2} A l , and suppose we want the inductor's length and diameter to be 1.8cm and 5.0mm respectively, and solving for n. The number of turns per meter, n, is (a) 324 turns/m. (a) 528 turns/m. (a) 324 turns/m.
37) The number of turns in the length of the indictor (in 1.8 cm length) is (a) 18 turns. (b) 9.5 turns. (c) 14.5 turns.