� AC CIRCUITS

• BRANCH-E & TC ENGG
• SUBJECT-ELECTRICAL MACHINE
• CHAPTER-4-AC CIRCUITS
• TOPIC-AC CIRCUITS
• SEM-4^TH
• FACULTY-Er.Gurupada Mishra (LECT. ELECTRICAL ENGG DEPARTMENT)
• AY-2021-2022, SUMMER-2022

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• Electricity supply systems are normally ac (alternating current).
• The supply voltage varies sinusoidal

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• instantaneous applied voltage,

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OR

where

• V_m = peak applied voltage in volts
• f = supply frequency in Hz
• t = time in seconds.

Introduction

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Instantaneous current,

Current and Voltage are in phase

i

Resistance connected to an AC supply

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• The “effective” values of voltage and current over the whole cycle

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• rms voltage is

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• rms current is

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Meters normally indicate rms quantities and this value is

equal to the DC value

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Other representations of Voltage or Current are

• maximum or peak value
• average value

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“RMS value of an alternating current is that steady state current (dc) which when flowing through the given resistor for a given amount of time produces the same amount of heat as produced by the alternative current when flowing through the same resistance for the same time”

Root Mean Square (rms) Voltage and Current

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i – instantaneous current

Current lags Voltage by 90 degree

rms current

Using complex numbers and the j operator

Inductive Reactance

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Inductance connected to an AC supply

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Phasor diagram and wave form

Current leads Voltage by 90 degrees

Capacitance Reactance

rms current

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i

Using complex numbers and the j operator

Capacitance connected to an AC supply

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Complex Impedance

Cartesian Form

-j indicates that the current lags the voltage

But

and

And

R and L in series with an AC supply

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-φ_L indicates lagging current.

In Polar Form

phasor diagram constructed with RMS quantities

Complex Impedance:

Cartesian Form:

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For the circuit shown below, calculate the rms current I & phase angle φ_L

Answer: I = 0.85A ∠-32.10

Exercise:

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but

Complex Impedance

The current, I in Cartesian form is given by

+j signifies that the current leads the voltage.

i

But

and

R and C in series with an AC supply

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In Polar Form

+φ_C identifies current leading voltage

phasor diagram drawn with RMS quantities

Power Factor

sinusoidal current leading the voltage

Complex Impedance:

I Cartesian form:

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For the circuit shown, calculate the rms current I & phase angle φ_L

Answer: I = 5.32mA ∠57.90

Exercise:

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Complex Impedance

But

&

V_C

V_L

V_R

We know that:

RLC in series with an AC supply

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The phasor diagram (and hence the waveforms) depend on the relative values of ωL and 1/ωC. Three cases must be considered

From previous page

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capacitive

resistive

inductive

Resonant frequency

From previous page

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From the above equation for the current it is clear that the magnitude of the current varies with ω (and hence frequency, f). This variation is shown in the graph

at ω_o,

• f_o is called the series resonant frequency.
• This phenomenon of series resonance is utilised in radio tuners.

and they may be greater than V

=

&

From previous page

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For circuit shown in figure, calculate the current and phase angle and power factor when frequency is

(i) 159.2Hz, (ii) 1592.Hz and (iii) 503.3Hz

(i) 11.04 mA + 83.6^o, 0.111 leading

(ii) 11.04mA, -83.6⁰, 0.111 lagging

(iii) 100mA, 00, 1.0 (in phase)

Answer:

How about you try this ?

Exercise:

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Can U name the Laws?

We know that:

and

Hence,

Substituting for the different Voltage components gives:

AC Supply in Parallel with C, and in Series R &L

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For the circuit shown calculate the minimum supply current, I_s and the corresponding capacitance C. Frequency is 50 Hz.

Answer: I_Smin = 3.71A C = 38.6μF

How about you try this one too?

☺

Exercise:

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power dissipation | _{instantaneous} = voltage| _{instantaneous} × current | _{instantaneous}

instantaneous voltage,

instantaneous current,

but

&

net power transfer

We know that:

Hence,

Therefore,

Power Dissipation

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i

P

P2

P1

Im

Re

V

P = Apparent power

P1 = Real power

P2 = Reactive power

θ

כ

Real, Apparent and Reactive Power

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i

P

P2

P1

Im

Re

O

V

P = Apparent power

P1 = Real power

P2 = Reactive power

P22

Pn

II

P22= New Reactive Power

Pn= New Apparent Power

I= Current to reduce Reactive Power

I

Power Factor Correction

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• Displacement transducers are often variable capacitors,
• Their capacitance varies with movement.
• The value may be adjusted by varying either
• the distance between the capacitance plates, or
• the effective plate area, or
• the effective dielectric between the plates

Capacitance

Where

ε₀ = permittivity of free space

ε_r = relative permittivity of dielectric

A = area of overlap between the plates

d = distance between the plates

Capacitance Transducers

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To determine the displacement by measuring the capacitance accurately. When the bridge is balanced,

To achieve the maximum bridge sensitivity:

• the two capacitors should be equal
• the resistances equal to the capacitive reactance at the measuring frequency.

For accurate measurements prevent or minimise:-

• stray capacitance between leads and earth
• transducer lead inductance
• transducer dielectric losses
• harmonic distortion (undesired components) in voltage supply

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Linearity of the transducer may be improved by using a

differentially connected displacement device

The transducer is connected to adjacent arms of an ac bridge.

Movement of the central plate increases the capacitance on one side and reduces it on the other.

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1. AC supply with resistive load, RL in series, RC in series, RLC in series, and RLC in parallel.
2. Phasor & Cartesian representations.
3. Phase angle and power factor.
4. Dissipated Power.
5. Applications: Capacitance transducer

Conclusion

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THANK YOU