Electrical- AC Circuits and Filters Flashcards
How to express alternating current in terms of peak current and phase
i=IpkSin(ωt+φ)
Where ω is angular frequency
φ is phase
How to express AC voltage in terms of peak voltage and phase
v=VpkSin(ωt+φ)
Where ω is angular frequency
φ is phase
What does a phaser look like?
An argand diagram with a half line representing an AC voltage or current. The length of half line is peak value and the angle between it and the positive real axis is its phase relative to the reference phase.
When a capacitor is in series with an AC voltage source, what do you do to find the current using the input voltage?
The capacitor scales the input voltage by ωC and shifts it’s phase by π/2 (-π/2 in sin function) to get current. Comes from i=Cdv/dt for transient circuit and v=VpkSin(ωt+φ)
Means current leads voltage
What does I=5<45° mean?
Peak current is 5A and the current starts 45° late.
What is a capacitor in an AC circuit with voltage of -jI/ωC equivalent to as a resistor?
A resistor with impedance -j/ωC. Because V=IR still applies
What is impedance?
Symbol Z. The complex equivalent of resistance. Consists of resistance (R) and reactance (X)
Z=R+Xj
Resistance loses energy whereas reactance stores and releases it
Which parts of impedance do capacitors and inductors have?
Only reactance not resistance. So only imaginary part of impedance.
When an inductor is in series with an AC current source, what do you do to find the voltage across it using the input current?
An inductor scales the input current by ωL and shifts its phase by π/2 (+π/2 in sin function) to get voltage. Comes from v=Ldi/dt for transient circuit and i=IpkSin(ωt+φ).
Means current lags voltage
What is an inductor with impedance ωLIj in series with an AC current source equivalent to as a resistor?
A resistor with impedance ωLj. Because V=IR still applies
General formula for RMS voltage
Vrms=sqrt( (1/T)Sv^2dt )
Where T is time period and integral is from 0 to T
S means integral
Formula for instantaneous power
p=(v^2)/R
Formula for average power for a resistor
P=(Vrms^2)/R
Derive the formula for Vrms for sinusoidal AC current
Start with Vrms^2=(1/T)Sv^2dt. Sub in 2π/ω for T and for v VpkSin(ωt+φ). Take out constant Vpk^2ω/2π. Use double angle formula to rewrite sin^2 function. Integrate between 2π/ω and 0 and some bits cancel (4π is 0). Expands and simplify to Vrms^2=(Vpk^2)/2. Square root both sides.
For a resistor and either capacitor or inductor in series in an AC circuit, why is the power dissipated across the both components not the Vrms x Irms across both?
Because the reactance (X) of the capacitor or inductor doesn’t dissipate power, only resistance does. This calculation gives apparent power (S).
What is the correct formula for power dissipated across a resistor and inducto/capacitor in series in an AC circuit when resistance is known?
P=Irms^2 x R
Or P=IVR (both rms)
Where R in VR is subscripts and means Vrms across the resistor
Or P=Irms^2 x Re(Z)
Where Z is combined impedance of both components
Derive formula for active power (P) in terms of Irms and Vrms for a resistor and inductor in series and AC circuit
Know P=Irms^2 x Re(Z).
Also V=IZ so Z=V/I=magnitude of Vrms/magnitude of Irms
Describe the power triangle
Right angled triangle with bottom line angle θ from hypotenuse.
Hypotenuse is apparent power, S (simple formula)
Adjacent is active power, P (with the power factor)
Opposite is reactive power, Q (mag(Irms^2) x X)
What do the 4 types of filter do?
Low-pass: voltages at low frequencies pass, voltages at high frequencies are blocked
High-pass: high frequency voltages pass
Band-pass: voltages between 2 frequencies pass
Band-stop: voltages between 2 frequencies are blocked
What is important to remember about the frequencies when finding if a certain frequency voltage passes?
The frequency of the voltage is not an angular frequency so must use ω=2πf
What is gain?
The ratio of power out to power in. Measured in decibels (dB).
Formula of Gain(dB)=10log(Po/Pi)
Where Po is power out
Pi is power in
Both are magnitudes
It is also the square of voltage gain so =20log(Vo/Vi)
On a Bode plot, what is the maximum gain?
0dB
What is the corner frequency?
The critical frequency of voltage in an AC circuit where the gain is about -3dB. It’s when the term in the logarithm for finding gain multiplied by angular frequency equals 1.
Steps for drawing a Bode plot for high or low pass filter
The axes are gain(dB) against ω (logarithmic scale).
- Find corner frequency and plot it at 0dB
- Draw horizontal line from point to ω=0 (low pass) or ω=infinity (high pass)
- Draw line with gradient -20dB/decade from point to x axis for low pass (+ and to point for high pass)
- The actual line curves under the plotted point so the corner frequency is at -3dB