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 voltagecurrent relation (Ohm's law) is:V_{rms} = R I_{rms } and V_{max} = R I_{max}.
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: V_{max} = 17.0 volts & ω = 377 rd/s.
I_{max} = V_{max} /R; I_{max} = 0.500A ; I_{rms} = 0.707I_{max} ; I_{rms} = 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 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 below. The unit of X_{c} 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 90^{o} when a capacitor is connected to an ac source.


Comparing the voltage and current graphs, note that when V_{C} is zero, I_{C} is maximum and when I_{C} is zero, V_{C }is maximum as expected. That is why there is a phase difference of 90^{o} 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: V_{max} = 8.50volts and ω = 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} = 0.170A ; I_{rms} = 0.707I_{max} ; I_{rms} = 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, X_{L}." Inductive reactance depends on two factors: one is the selfinductance 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 X_{L} 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 90^{o} 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 V_{L} is zero, I_{L} is maximum and when I_{L} is zero, V_{L}_{ }is maximum as expected. That is why there is a phase difference of 90^{o} between the two.
Example 3: In the above figure, suppose V = 36sin(720t) and L = 250mH. Find the rms current.
Solution: From the equation: V_{max} = 36 volts and ω =720 rd/s.
X_{L} = Lω ; X_{L} = (0.25H)(720 rd/s) = 180 Ω
I_{max} = V_{max}/X_{L} ; I_{max} = 0.20A. ; I_{rms} = 0.707I_{max} ; I_{rms} = 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 I_{max} = V_{max}/Z & 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(400t). Find the I_{rms} and the phase angle φ.
Solution: From the equation: V_{max} = 18.0 volts and ω = 400rd/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} = 0.600A ; I_{rms} = 0.707I_{max} ; I_{rms} = 0.424A.
φ = tan^{1} [ (X_{L} X_{C })/R ] = tan^{1}[0.353] = 19.4^{o}.
LC Resonance Circuits:
If R = 0 in an RCL 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 capacitance 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) 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^{9}F)(4.0x10^{6}H)] = 1.0x10^{6} rd/s.
(b) ω = 2π f ; f = w/2π ; f = 1.0x10^{6}/(2π) = 160,000Hz or 160kHz.
(c) T = 1/f ; 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 L_{1}L_{2} with another inductor that receives signals from an antenna. Inductor L_{1} in the antenna circuit induces a current in 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. 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 0.050pFcapacitor is used, find the inductance L of the inductor to be used to create resonance at that frequency.
Solution: (a) ω = 2πf ; ω = 2π(2.0x10^{9}Hz) = 1.3x10^{10}rd/s.
(b) ω = 1/(LC)^{1/2} ; ω^{2} = 1/LC ; L = 1/[Cw^{2}] = 1.27x10^{7}H
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: The formula for selfinductance L is: 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 = 507turns/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 90^{o} (c) I lags V by 90^{o}.
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 90^{o} (b) leads the voltage by 90^{o} (c) leads the voltage by 45^{o}. 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 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 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 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 in an AC source, current (a) leads voltage by 90^{o} (b) leads voltage by 45^{o} (c) lags voltage by 90^{o}.
16) The inductive reactance X_{L} 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 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 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 X_{C} is (a) 330Ω (b) 60Ω (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) 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) X_{C}  X_{L} = 0 (b) Lω = 1/(Cω) (c) both a & b.
33) In Question 32, solving for ω by crossmultiplication, yields: (a) LCω^{2} = 1 (b) ω^{2} = 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) 1.16x10^{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 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.