- An ac voltage source has a sinusoidally varying output: v = V
_{m}sin 2p f t - RMS voltage and current: V
_{r}= v_{m}/ sqrt(2), I_{r}= I_{m}/ sqrt(2) - See the summary after section 21.3.

And finally, consider what happens when an inductor is put in a circuit with an ac voltage source. Similar arguments apply as for the capacitor, but this time we find that the current leads the voltage by a quarter cycle (90°), and that the rms current is related to the rms voltage across the solenoid by:

V_{L} = I_{r} X_{L}

The quantity X
X_{L} = 2p f L

A 2.40mF capacitor is connected across an alternating voltage with an rms value of 9.00V. The rms current in the capacitor is 25.0mA. (a) What is the source frequency? (b) If the capacitor is replaced by an ideal coil with an inductance of 0.160H, what is the rms current in the coil?

(a) From the rms voltage and current, we can determine the capacitive reactance, and this is related to the capacitance and the frequency of the source.

X_{C} = V_{r} / I_{r} = 9.00V / 25.0mA = 360W.

f = 1 / (2p C X_{C}) = 1 / (2p(2.40mF)(360W)) = 184Hz.

At this frequency, the inductive reactance is

X_{L} = 2p f L = 2p (184Hz)(0.160H) = 185W.

The rms current can be found from the rms voltage and reactance:

I_{r} = V_{r} / X_{L} = 9.00V / 185W = 48.6mA.

Resistor | Capacitor | Inductor | |
---|---|---|---|

units | ohm, W = V / A | farad, F = C / V | henry, H = V s / A |

symbol | R | C | L |

reactance | X_{R} = R |
X_{C} = 1 / (2p f C) |
X_{L} = 2p f L |

phase | 0° | -90° | 90° |

Now we want to examine what happens when a resistor, capacitor, and inductor are combined in an ac circuit.
We begin by considering the circuit consisting of an ac voltage source, a resistor, a capacitor, and an inductor in series (sketch circuit like Figure 21.8).
What should we expect to happen in this circuit, that is, what will be the relation between voltage and current?
This is a series circuit, so the current will be the same through all the elements.
Although the current is the same, the voltages across each element will in general be different in magnitude and in *phase*.
The result is that the voltage from the ac source need not be in phase with the current, and the relative phase can be essentially anything.

Now let's work this out quantitatively. If we let the current be

i = I_{m} sin 2p f t

then the voltage drop across the resistor is
DV_{R} = i R,

the voltage drop across the capacitor is
DV_{C} = i X_{C},

and the voltage drop across the inductor is
DV_{L} = i X_{L}.

Note that the frequency of the current is the same as the frequency of the voltage source.
From the previous sections we know that the voltage drop across the resistor is in phase with the current, the voltage drop across the capacitor lags the current by 90°, and the voltage drop across the inductor leads the current by 90°.
We can represent all of this with a
DV = Sqrt{DV_{R}² + (DV_{L} - DV_{C})²}

tan f = (DV_{L} - DV_{C}) / DV_{R}

Because the voltages are related to the reactances by a common factor, the current, we can also make a phasor diagram for the reactances. The hypotenuse of the reactance diagram is called the impedance,

Z = Sqrt(R² + (X_{L} - X_{C})²}.

tan f = (X_{L} - X_{C}) / R

In terms of the impedance, we can write Ohm's law for ac circuits:
DV = I Z

where V and I can be either rms values, or peak values, but don't mix them and choose one of each!
A 50.0W resistor, a 0.100H inductor, and a 10.0mF capacitor are connected in series to a 60.0Hz source. The rms current in the circuit is 2.75A. Find the rms voltages across (a) the resistor, (b) the inductor, (c) the capacitor, and (d) the RLC combination. (e) Sketch the phasor diagram for this circuit.

Since the rms voltage is related to the peak voltage by a constant, we can use the Ohm's law relations for rms voltages and currents.

(a) V_{R}(rms) = I_{r} R = (2.75A)(50.0W) = 138V_{rms}.

(b) X_{L} = 2p f L = 2p (60.0Hz)(0.100H) = 37.7W

V_{L}(rms) = I_{r} X_{L} = (2.75A)(37.7W) = 104V_{rms}.

(c) X_{C} = 1/2p f C = 1 / (2p(60.0Hz)(10.0mF) = 265W

V_{C}(rms) = I_{r} X_{C} = (2.75A)(265W) = 729V_{rms}

(d) The rms voltage drop across the RLC combination can be determined using the expression for the hypotenuse of the voltage triangle.

DV = Sqrt{V_{R}² + (V_{L} - V_{C})²} = Sqrt{(138V)² + (104V - 729V)²} = 640V.

As mentioned previously (see the summary table at the beginning of the lecture of Oct. 20, 2000), capacitors and inductors store energy, but do not dissipate energy. Only resistors dissipate energy. Therefore, the power dissipated in an ac circuit depends only on the resistance in the circuit

P_{av} = I² R.

Since R = DV
P_{av} = I DV_{R}.

Finally, we can express DV
P_{av} = I DV cosf.
The cos f factor is known as the power factor.

In a certain RLC circuit, the rms current is 6.0A, the rms voltage is 240V, and the current leads the voltage by 53°.
(a) What is the total resistance of the circuit?
(b) Calculate the total reactance, X_{L} - X_{C}.
(c) Find the average power dissipated in the circuit.

Begin by drawing a phasor diagram of the rms voltage.
Because the current leads the voltage, the voltage lags the current, so the phase angle is f = -53°.
(a) V_{R} is the component of the voltage along the x-axis, and is related to the resistance by R = V_{R} / I_{r}.

V_{R} = DV cosf = (240V) cos-53° = 144V

R = V_{R} / I_{r} = (144V) / (6.0A) = 24W

(b)X_{L} - X_{C} = R tanf = 24W tan-53° = -32W

(c) P_{av} = I V cosf = (6.0A)(240V)cos-53° = 870 W.