Electrical- AC Circuits and Filters

Question 1

How to express alternating current in terms of peak current and phase

Answer 1

i=IpkSin(ωt+φ)
Where ω is angular frequency
φ is phase

Question 2

How to express AC voltage in terms of peak voltage and phase

Answer 2

v=VpkSin(ωt+φ)
Where ω is angular frequency
φ is phase

Question 3

What does a phaser look like?

Answer 3

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.

Question 4

When a capacitor is in series with an AC voltage source, what do you do to find the current using the input voltage?

Answer 4

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

Question 5

What does I=5<45° mean?

Answer 5

Peak current is 5A and the current starts 45° late.

Question 6

What is a capacitor in an AC circuit with voltage of -jI/ωC equivalent to as a resistor?

Answer 6

A resistor with impedance -j/ωC. Because V=IR still applies

Question 7

What is impedance?

Answer 7

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

Question 8

Which parts of impedance do capacitors and inductors have?

Answer 8

Only reactance not resistance. So only imaginary part of impedance.

Question 9

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?

Answer 9

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

Question 10

What is an inductor with impedance ωLIj in series with an AC current source equivalent to as a resistor?

Answer 10

A resistor with impedance ωLj. Because V=IR still applies

Question 11

General formula for RMS voltage

Answer 11

Vrms=sqrt( (1/T)Sv^2dt )
Where T is time period and integral is from 0 to T
S means integral

Question 12

Formula for instantaneous power

Answer 12

p=(v^2)/R

Question 13

Formula for average power for a resistor

Answer 13

P=(Vrms^2)/R

Question 14

Derive the formula for Vrms for sinusoidal AC current

Answer 14

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.

Question 15

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?

Answer 15

Because the reactance (X) of the capacitor or inductor doesn’t dissipate power, only resistance does. This calculation gives apparent power (S).

Question 16

What is the correct formula for power dissipated across a resistor and inducto/capacitor in series in an AC circuit when resistance is known?

Answer 16

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

Question 17

Derive formula for active power (P) in terms of Irms and Vrms for a resistor and inductor in series and AC circuit

Answer 17

Know P=Irms^2 x Re(Z).

Also V=IZ so Z=V/I=magnitude of Vrms/magnitude of Irms

Question 18

Describe the power triangle

Answer 18

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)

Question 19

What do the 4 types of filter do?

Answer 19

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

Question 20

What is important to remember about the frequencies when finding if a certain frequency voltage passes?

Answer 20

The frequency of the voltage is not an angular frequency so must use ω=2πf

Question 21

What is gain?

Answer 21

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)

Question 22

On a Bode plot, what is the maximum gain?

Answer 22

0dB

Question 23

What is the corner frequency?

Answer 23

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.

Question 24

Steps for drawing a Bode plot for high or low pass filter

Answer 24

The axes are gain(dB) against ω (logarithmic scale).

1. Find corner frequency and plot it at 0dB
2. Draw horizontal line from point to ω=0 (low pass) or ω=infinity (high pass)
3. Draw line with gradient -20dB/decade from point to x axis for low pass (+ and to point for high pass)
4. The actual line curves under the plotted point so the corner frequency is at -3dB

Question 25

Describe an example of a series resonant circuit

Answer 25

All in series. AC voltage source, capacitor, inductor, resistor. Vout is measured across resistor.

Question 26

How to get equation for gain for series resonant circuit

Answer 26

Treat inductor and capacitor as like one component and do potential divider formula. Get Vout=VinR/(R+j(ωL-1/ωC)). Rearrange to get voltage gain. Apply magnitude rules (square terms then root them). Get Gain(dB)=20log(R/sqrt(R^2+(ωL-1/ωC)^2)).

Question 27

How does the series resonant circuit act as a band-pass filter?

Answer 27

At low frequencies, 1/ωC&raquo_space;1 so gain is small
At high frequencies, ωL&raquo_space;1 so gain is small
At the resonant frequency ωL=1/ωC so the gain is 0dB. This frequency is therefore 1/sqrt(LC) and the overall impedance at this frequency is just R. Bode plot looks like a hill (broad or sharp peak). The pass band is the area under the shape above gain=-3dB

Question 28

What is the band width for a band-pass filter and specifically a series RLC circuit?

Answer 28

Band the width is the width of the pass band on the Bode plot in Hz.
For series RLC circuit it is R/L

Question 29

What is quality factor?

Answer 29

A measure of the ‘peakiness’ of the gain response for a filter. Defined by the equation
q=resonant frequency/band width
It is also the magnification of the voltage or current

Question 30

What is damping factor?

Answer 30

Defined by the equation
ζ=2/q
Where q is quality factor

Question 31

Describe a parallel resonant circuit

Answer 31

An AC voltage source in series with a resistor and a separate branch containing an inductor and capacitor in parallel with each other.

Question 32

How does the parallel resonant circuit act as a band stop filter?

Answer 32

At some frequency, the reactance of inductor will equal the negative of the reactance of the capacitor. The total current going into the parallel bit will equal the voltage on inductor/jXL + voltage on capacitor/jXC. At this critical frequency, the two terms equivalent to total current equal the negative of each other so cancel and total current is 0 in the circuit. The resonant frequency is 1/sqrt(LC). Bode plot has horizontal bits either side of resonant frequency and both lines go in to tend to -infinity gain

Question 33

What is the quality factor of a parallel RLC circuit?

Answer 33

Rsqrt(C/L)

One over what it is for series

Question 34

Examples of high pass filters

Answer 34

Series RC circuit where vout is across resistor

Series LC circuit where vout is across inductor

Question 35

What does graph of phase against frequency look like for a high or low pass filter?

Answer 35

Phase in degrees or radians on linear y axis. Angular frequency on logarithmic x axis. The frequencies where there is unity gain, there is horizontal line at phase=0. The frequencies where there is roll-off gain, there is a horizontal line at phase=90°. The middle ends of these lines are equidistant from the corner frequency which is at 45° phase. There is a curvy line from each end through the corner frequency in a shape like a sideways tan graph from -90 to 90.

Question 36

For a moving system with forces involved, how would you construct an analogous circuit?

Answer 36

Equate the forces using netF=ma. Write all accelerations, velocities and displacements in terms of velocity using integrals and derivatives. Say velocity is current. Match expressions in the equation to components for an AC circuit (di/dt is inductor, i is resistor, Sidt is capacitor). Say the expressions are potential differences across these components and draw the circuit so that the signs in the motion equation match those from voltage law equation.

Question 37

What can analogous circuits be used to find?

Answer 37

Natural frequency of oscillating system. Damping factor of oscillating system.