Transcript Inductors in an AC Circuit

• Means of power for many appliances we rely on to work in our homes, schools,
offices, etc.
• Circuits put together all components we have previously seen including resistors,
inductors, and capacitors and are driven by a sinusoidal voltage.
• What can we get from these circuits?
• The amplitude and time characteristics of the alternating current
• Discussion concluded with transformers, power transmission, and electrical filters
• Elements of AC circuit elements along with the power source provide and alternating
voltage (Δν) described by:
Δν=Δvmaxsinωt
• Sources include: electric outlets and generators
• Due to the sinusoidal properties of the voltage with time, the voltage is positive
during half of the cycle and negative during the other half
• The angular frequency for the AC voltage is:
ω = 2πƒ = 2π/T
ƒ = frequency
T = period
• The source determines the frequency of the current connected to it
• Considering a circuit with a resistor:
• At any instant, the algebraic sum of the voltages around a closed loop in a circuit must
be zero by Kirchhoff’s loop rule, so the voltage across the resistor is:
Δν – iRR = 0
• To find the instantaneous current in the resistor substitute Δvmaxsinωt for Δν:
iR = Δν/R = (Δvmax/R) sinωt = Imaxsinωt
• Imax is the maximum current: Imax=ΔVmax/R
• The instantaneous voltage across a resistor is:
ΔνR=iRR=ImaxRsinωt
• The average current in an AC circuit is important to determine
• We call this average current the rms current and relate it to the Imax by:
Irms = Imax/√2
also related to voltage by: ΔVrms = Δvmax/√2
• The average power delivered to a resistor that carries and alternating current is:
Pavg = I2rmsR
• The voltage we receive to power our appliances from a wall outlet is 120 V. Because
this source is an AC current, the 120 V is described as an rms value.
• Considering a circuit with and inductor:
• Kirchhoff’s rule for and inductor in a loop circuit gives:
Δν – L(diL/dt) = 0
• To find the instantaneous current in the inductor substitute Δvmaxsinωt for Δν:
Δν = L(diL/dt) = Δvmaxsinωt
• Solving for diL gives: diL =( ΔVmax/L)sinωt dt
• Integrating this expression give the instantaneous current iL in the inductor as a
function of time:
iL = (Δvmax/L) sinωt dt = -(ΔVmax/ωL) cos ωt
• The previous equation can be expressed differently by substituting cos ωt = -sin (ωt –
π/2):
iL = (Δvmax/ωL) sin (ωt – π/2)
• The instantaneous current in the inductor and the instantaneous voltage across the
inductor are out of phase by π/2 radians. The plot shows:
• The current in an inductive circuit reaches its maximum value when cosωt=±1:
Imax=Δvmax/ωL
• ωL behaves the same as resistance in a current so it is given the same units as
resistance (Ω)
• We define ωL as inductive reactance (XL):
XL=ωL
• Therefore we can write:
Imax=Δvmax/XL
• The rms value for an inductor is found by:
Irms=Δvrms/XL
• The instantaneous voltage across and inductor is:
ΔνL = -L(diL/dt) = -Δvmaxsinωt = -ImaxXLsinωt
• Considering a circuit with and capacitor:
• Kirchhoff’s rule for and capacitor in a loop circuit gives:
Δν – (q/C) = 0
• To find the instantaneous current in the capacitor substitute Δvmaxsinωt for Δν:
q = CΔVmaxsinωt
• The instantaneous current in the circuit is:
iC = dq/dt = ωCΔVmaxcosωt
• Using cosωt = sin (ωt + π/2), we get:
iC = ωCΔVmaxsin(ωt + π/2)
• The current is also π/2 radians out of phase with the voltage across a capacitor
• The current in an capacitive circuit reaches its maximum value when cosωt=±1:
Imax = ωCΔVmax = ΔVmax/(1/ωC)
• 1/ωC behaves the same as resistance in a current so it is given the same units as
resistance (Ω)
• We define 1/ωC as capacitive reactance (Xc):
Xc=1/ωC
• Therefore we can write:
Imax=Δvmax/XC
• The rms value for an inductor is found by:
Irms=Δvrms/XC
• The instantaneous voltage across and inductor is:
ΔνC = Δvmaxsinωt = ImaxXCsinωt
• As frequency of the voltage source increases, the capacitive reactance decreases and
the maximum current increases
• We have considered all the parts separately, but now we will see how the circuit act
with the resistor, inductor, and capacitor in a seires
• The instantaneous voltages across the three circuit elements are:
ΔνR = ImaxRsinωt = ΔVRsinωt
ΔνL = ImaxXLsin(ωt + π/2) = ΔVLcosωt
ΔνC = ImaxXCsin(ωt + π/2) = -ΔVCcosωt
• The sum of these three voltages must equal the voltage from the AC source, but they
cannot be added directly because the three voltages have different phase
relationships with the current.
• We can relate the voltages by:
Δvmax = √[ΔVR2 + (ΔVL – ΔVC)2 = √[(ImaxR)2 + (ImaxXL – ImaxXC)2]
Δvmax = Imax √(R2 + (XL – XC)2]
• We can then express the maximum current as:
Imax= Δvmax/√(R2 + (XL – XC)2]
• The denominator of the fraction Imax= Δvmax/√(R2 + (XL – XC)2] plays a role in resistance
and is called impedance (Z) of a circuit:
Z = √(R2 + (XL – XC)2
• Impedance also has the units of ohms just as resistance. We can rewirte the equation
for maximum current to be:
Imax = Δvmax/Z
• The phase angle Φ between the current and the voltage is found as:
Φ = tan-1[(ΔVL-ΔVC)/ΔVR] = tan-1[(ImaxXL – ImaxXC)/ImanR]
So, Φ = tan-1[(XL-XC)/R]
• The instantaneous power delivered by an AC source to a circuit is the product of the
current and the applied voltage.
• For the RLC circuit, we can express the instantaneous power (P) as:
P = iΔν = Imaxsin(ωt – Φ) Δvmaxsinωt
P = Imax Δvmaxsinωt sin(ωt-Φ)
• This expression leads to a complicated function of time so we will look at power as an
average over one or more cycles. Therefore we will use:
P = Imax Δvmax sin2 ωt cos Φ – Imax Δvmax sinωt cosωt sinΦ
• Removing the constants, we can simplify this expression to express average power as:
Pavg = ½Imax Δvmax cosΦ
• Expressing average power in terms of the rms current and rms voltage gives:
Pavg = Irms ΔVrms cosΦ or Pavg = I2rmsR or Pavg = Irms ΔVrms
• An RLC circuit is an electrical oscillating system.
• The rms current is written:
Irms = ΔVrms/Z
Z = impedance
• Substituting for Z we get:
Irms = Δvrms/√(R2 + (XL – XC)2]
• Because the impedance depends on the frequency of the source, the current in the
RLC circuit also depends on the frequency. The angular frequency (ω0) at which XLXC=0 is called the resonance frequency of the circuit.
• To find ω0, we set XL=XC, which gives:
ω0 = 1/√(LC)
• The power traveling through power lines is delivered at a high voltage around 20,000
V. There must be a way to decrease the voltage to the 120 V that is delivered by our
power outlets.
• AC Transformers consists of two coils of wire would around a core of iron are
responsible for this decrease in voltage from the power delivery station to our homes.
• The coil on the left is considered the primary coil because it is connected to the input
alternating-voltage source and has N1 turns. The coil on the right consisting of N2
turns is called the secondary winding.
• The iron core increases the magnetic flux through the coil to provide a medium in
which nearly all the magnetic field lines through one coil pass through the other coil.
• Using transformers, Faraday’s law states that the voltage
Δν1 = -N1 (dΦB/dt) where,
ΦB is the magnetic flux through each turn
• If we assume all magnetic field lines remain within the iron core, the flux through
each turn of the primary equals the flux through each turn of the secondary.
Therefore the voltage across the secondary is:
Δν2 = -N2 (dΦB/dt)
• Solving for dΦB/dt and substituting the result into Δν2 = -N2 (dΦB/dt) we get:
Δν2 = (N2/N1) Δν1
• Filters can be designed to respond to different frequencies
• The input voltage is across the series combination of the two elements.
• The output is the voltage across the resistor
• At low frequencies, Δvout is much smaller than Δvin, whereas at high frequencies, the
two voltages are equal
• Because the circuit preferentially passes signals of higher frequency while blocking
low –frequency signals, the circuit is called an RC high-pass filter
• The capacitor blocks out direct current and AC current at low frequencies
• Filters can be designed to respond to different frequencies
• The capacitor and resistors positions have been swapped in this filter so that the
output voltage is taken across the capacitor
• At low frequencies, the reactance of the capacitor and the voltage across the
capacitor is high
• As the frequency increases, the voltage across the capacitor drops
• This filter is an RC low-pass filter