PS114
Chapter 23
Alternating Current Circuits
1
Capacitors and capacitive reactance
• In this chapter we introduce other electrical elements into the ac circuit,
namely the capcitance C and the inductance L
• Recall the relationship for the voltage across a resistor in an ac circuit:
VR = Irms R
• In an ac circuit containing just a capacitor C, we can write the voltage
across the capacitor as:
VC = Irms XC
(similar to Ohm’s law)
where XC is called the capacitive reactance.
XC =
Problem 2:
1
1
=
ωC
2πf C
[ohms]
What voltage is needed to create a current of 35 mA in a circuit
containing only a 0.86-µF capacitor, when the frequency is 3.4
kHz?
1
2
Inductors and inductive reactance
• In an ac circuit containing just an inductor L, we can write the voltage
across the inductor as:
VL = Irms XL
(similar to Ohm’s law)
where XL is called the inductive reactance.
XL = ωL = 2πf L
Problem 8:
[ohms]
The current in an inductor is 0.20 A, and the frequency is 750 Hz.
If the inductance is 0.080 H, what is the voltage across the inductor?
2
3
Circuits containing resistance, capacitance, and inductance
• The peak voltage for an ac circuit containing R, L, and C is:
Vo2 = VR2 + (VL − VC )2
where VR , VL , and VC represent the peak voltages.
Likewise, one can also write the rms voltage as:
2
Vrms
= VR2 + (VL − VC )2
where VR , VL , and VC represent rms quantities.
• Finally, we can write one equation for Ohm’s law for an RLC circuit:
q
Vrms = Irms R2 + (XL − XC )2
Vrms = Irms Z
Ohm’s Law
q
Z = R2 + (XL − XC )2
the “impedance”
• The phase angle φ is defined as angle between the resultant voltage V
and the
current I:
VL − VC
XL − XC
tan φ =
=
VR
R
• The average power delivered by the ac source is:
2
P = Irms
R
However, cos φ = VR /Vo = (Irms R)/(Irms Z) = R/Z, so R = Z cos φ.
Now we can write the average power as:
2
P = Irms
Z cos φ = Irms (Irms Z) cos φ
P = Irms Vrms cos φ
(the average power in an RLC circuit)
3
Problem 16:
A light bulb has a resistance of 240 Ω. It is connected to a
standard wall socket (120 V, 60.0 Hz). (a) Determine the current
in the bulb. (b) Determine the current in the bulb after a 10.0µF capacitor is added in series in the circuit. (c) It is possible to
return the current in the bulb to the value calculated in part (a)
by adding an inductor in series with the bulb and the capacitor.
What is the value of the inductance of this inductor?
4
4
Resonance in RLC circuits
• In an RLC circuit driven by an ac power source, it is possible to transmit large amounts of energy between the source and the circuit if the
resistance is low, and the driving frequency f is close to the natural
frequency of oscillation of the RLC circuit.
• The natural oscillating frequency of an RLC circuit can be found by
setting XL = XC . By doing so, this makes Z as small as possible, and
permits the largest possible Irms to be transfered into the circuit because
Irms = Vrms /Z.
• By setting XL = XC , we can find the natural frequency of oscillation of
the LC circuit.
XL = XC
⇒
ωL =
1
ωC
⇒
ωo2 =
1
LC
⇒ (2πfo )2 =
1
LC
Finally, we can write the resonant oscillation frequency fo as:
fo =
1
√
2π LC
where fo is in units of Hz.
Problem 26:
5
A series RLC circuit has a resonant frequency of 690 kHz. If the
value of the capacitance is 2.0 × 10−9 F, what is the value of the
inductance?
Semiconductor devices
• This section is worth reading, but there are no homework
problems for this section.
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