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Thursday, 22 August 2019

Wien Bridge and Resonance Bridge

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Wien Bridge:


Wien bridge is primarily popular as a frequency determining bridge. It has several applications. Some of the applications are in harmonic distortion analyser, in audio frequency generation. It has a series RC combination in the adjoining arm. The other two arms are provided with resistances. The impedance of arm 1 is Z1 = R1 – j/ωC1. The admittance of arm 3, is Y3 = 1/R3 + j ωC3. The arrangement is shown in figure.


The balance equation can be obtained as follows:



R2 = (R1 – j/ωC1) R4(1/R3+jωC3) ------------------------- 1

Expanding the expression we have:

R2 = R1R4/R3 + (jωC3R1R4) – jR4/ ωC1R3 + R4C3/C1------------------------- 2

Equating the real terms:

R2 = R1R4/R3 + R4C3/C1 ------------------------- 3

It can be reduced to:

R2/R4 = R1/R3 +C3/C1 ------------------------- 4

Equating the imaginary terms:

ωC3R1R4 = R4/ ωC1R3 ------------------------- 5

where, ω = 2πf

Solving for f, we get

f = 1/2π ( C1C3 R1R3) ------------------------- 6

From the above we observe that the two conditions for balance result in an expression determining the required resistance ratio, R2/R4 and another expression determining the frequency of the applied voltage. So we have to satisfy the above equation (4) and (6), to balance the bridge.

The arrangement in most of the wien bridge circuits is such that the values of R1, R2 and C1,C3 are made equal. Hence the equation (4) reduces to R2/R4 = 2. It also reduces equation (6) to

f = 1/2 πCR ------------------------ 7

Thus Equation (7) is the general expression for the frequency of the Wien Bridge.
Practically Capacitors C1 and C3 are fixed values. R1 and R3 are variable resistors controlled by a common shaft. Now providing R2 = 2R4, the bridge can be used as a frequency determining device with single balance control, which can be calibrated directly in terms of frequency. The source supplying this bridge must be free from harmonics. If not, the balancing will be difficult. Hence it is clear that the bridge is frequency sensitive.

Resonance Bridge:


Resonance bridge consists of reactance concentrated in one arm. They are adjusted to give series resonance so that this arm offers resistance impedance. The resonance bridge is shown in figure. From the schematic diagram of the bridge, we find that the ratio arms are formed by R1 and R2. Resistance R3 is connected in the standard arm. The fourth arm consists of an inductance Lx, capacitance Cx and resistance Rx.

This bridge can be used to measure frequency in terms of inductance and capacitance. It is also used to measure capacitance in terms of frequency and a variable inductance. It can also be used to measure inductance in terms of frequency and a variable capacitance. The balance equation can be obtained as follows:

R1(Rx + jωLx – j/ωCx) = R2 R3

At resonance

XL = Xc and fx = 1/2π (LC)

Zx = Rx

Therefore, Rx = R2 R3/R1

As can be seen from the above the bridge is balanced by resistance alone. Resistance R3 is used for this purpose.


Wednesday, 21 August 2019

Why single phase induction motor is not self starting?

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Induction motor:

Induction motor is a machine which converts electrical energy into mechanical energy. There is no electrical connection between the supply voltage and the motor winding. E .m. f is induced in the rotor by electromagnetic induction. So it is called induction motor.
Induction motor are classified into two types. They are,

(1). Single phase Induction Motor
(2). Three phase induction motor

SINGLE PHASE (1Φ) INDUCTION MOTOR

It is motor with only one set of winding wound on the stator. It has a squired cage rotor. There is uniform air gap between stator and rotor. Stator and rotor are not electrically connected.

Why single phase induction motor is not self starting?

In single phase (1Φ) induction motor, the rotor is a squirrel cage rotor. In stator, only one winding (main winding) is provided. When an a.c supply (single phase) is given to the stator winding, an alternating (pulsating) magnetic field is produced. Therefore the rotor rotates in both directions alternatively. But the rotor does not rotate continuously. So the single phase (1Φ) induction motor is not self starting.

The following method is used to self start a single phase (1Φ) induction motor

1. Phase splitting method

(a) By using starting winding
(b) By using a capacitor in the winding
(c) By shaded poles.

2. Series motor method.

3. Repulsion method.

Types of single phase (1Φ) induction motor

1. Split phase induction motor
2. Capacitor start induction motor
3. Shaded pole induction motor
4. Repulsion motor

Transformer Losses-Regulation-Efficiency Calculation

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Losses in a Transformer is divided into two types,

1. Core (or) iron losses
2. Copper losses

1. Core (or) iron losses

These losses consist of hysteresis and eddy current losses and occur is the transformer core due to the alternating flux. These losses can be determined for open circuit.

Hysteresis loss α f Bm1.6
Eddy current loss α f2 Bm2 t2

All of the hysteresis and eddy current losses depend upon maximum flux density (Bm) in the core frequency. Due to transformers are merged to constant frequency, constant voltage, f and Bm are constant. Hence it losses are same at all loads.



Core (or) iron loss, Pi = hysteresis loss eddy current loss = constant losses

2. Copper losses

These losses occur in both the primary and secondary windings due to ohmic resistance. These losses can be determined from short circuit test.

Primary winding copper loss = I12R1
Secondary winding copper loss = I22R2
Total copper losses, Pc = I12R1 + I22R2
= I12R01 (or) I22R02

Copper loss occurs when the current flows through the winding. It is equal to I2R. The loss varies as square of the load current. Since the copper losses vary with load current it is called variable loss.

Total losses in a transformer = constant loss + variable loss
= Pi + Pc

Efficiency of a transformer

Efficiency of a transformer is defined as the ratio of output power to input power

% efficiency , ή = 100 × Out power/ Input power

Efficiency of a transformer can be calculated, after determine the losses

= Output power/ (Output power + losses)
= Output power/ (Output power + copper losses + iron losses)

All day efficiency

The all day efficiency is defined as the ratio of output is kWh to the input is kWh of a transformer during the whole day.
All day efficiency = Output is kWh/ Input is kWh for 24 hrs

Condition for maximum efficiency

The condition for maximum efficiency is copper loss should be equal to iron loss.
Condition of maximum efficiency is,
Copper loss = iron loss

Efficiency , ή = Output power / Input power

ή = Output power/ Output power + losses

Losses = iron loss + copper loss

Output power = V2I2 cosθ2

Core loss, pi = Hysteresis loss + Eddy current loss
Copper loss pc = I12R1 + I22R2 = I22R02

Efficiency , ή = V2I2 cosθ2 / (V2I2 cosθ2+Pi+ I22R02)

Divided both numerator and denominater by I2

For maximum efficiency, denominator should be minimum I2

d/d I2 ( V2 cosθ2 + Pi / I2+I2R02) = 0
0 - Pi / I22 + R02 = 0
R02 = Pi / I22
Pi = I2R02

Core (or) iron loss = copper loss
This is the condition for maximum efficiency.

Regulation of a Transformer

When the transformer secondary is connected to the load, the current flow through the secondary winding resistance and reactance causing a voltage drop. This voltage drop increases with increase in current. Therefore terminal voltage across at the secondary winding changes. The change in voltage from no load to full load is known as the regulation of the transformer.

% regulation = No load voltage - full load voltage / No load voltage
% regulation = 100 × (E-V)/E
Where, E = no load voltage
V = full load voltage

Application of transformer

1. The transformer is used to either step up or step down the voltage
2. It is used in power supply for electronic circuit
3. In generating station the transformer step up the voltage for transmission.
4. It is used as an auto transformer starter for starting the induction motor.
5. It is used as an instrument transformer for increasing the range of meters.

Monday, 19 August 2019

Series and Parallel Resonance in AC Circuits

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Resonance in AC Circuits

An AC circuit is said to be in resonance when the circuit power factor is unity. i.e., At XL = XC resonance occurs. In a resonant circuit, resistors, Inductors and capacitors are connected either in series or in parallel combination. This circuit acts as a pure resistive circuit.

The frequency response of a resonant circuit is shown in the figure.

The frequency at which resonance occurs is called the resonant frequency (fr). From fig, the frequency response decreases sharply on either side of the resonant frequency. Hence a resonant circuit selects a range of frequencies for which the response is nearly equal to the maximum.



Two types of resonance are,

1. Series resonance
2. Parallel resonance

SERIES RESONANCE IN AC CIRCUITS

In a series resonance circuit, resistor (R), inductor (L) and capacitor(C) are connected in series across a sinusoidal voltage of variable frequency source. An R-L-C series resonance circuit is shown in the fig (a).

Resonant frequency

At resonance XL = XC

Phasor Diagram

Therefore, VL and VC are equal in magnitude but opposite in direction, hence they cancel each other.

Applied voltage, V = VR

Impedance is given by,

At resonance, the circuit current is maximum and the impedance is minimum. This is shown in the figure.

Power factor

cosθ = R/Z = R/R = 1(unity)

Quality factor of a series resonant circuit

At series resonance, the voltage across L and C is greater than the applied voltage. This voltage magnification is called Q-factor of the series resonant circuit.

Resonance curve of a series resonant circuit

The curve draw between current and frequency is known as resonance curve. Figure shows the resonance curve of a typical R-L-C series circuit.

At resonant frequency (fr), the inductive reactance XL is equal to the capacitive reactance XC. The impedance of the circuit is only resistive and equal to R. hence the current at resonance is maximum and the power factor is unity.

If the frequency of the circuit is less than resonant frequency, then XC>XL and the circuit behaves as a R-C circuit. The current is decreased due to increased impedance, and the power factor is leading.

If the frequency of the circuit is greater than resonant frequency, then XL>XC and the circuit behaves as a R-L circuit. The current is also decreased and the power factor is lagging.

Bandwidth of a series resonant circuit
The current versus frequency curve for a series resonant circuit is shown in the figure. At resonance, the current is maximum.

Bandwidth of a circuit is defined as the frequencies which lie between two points on either side of the resonant frequency where current is 0.707 of its maximum value.
As shown in the fig,

Bandwidth, BW = f2 - f1

Where, f1 = Lower cut-off frequency.
f2 = Upper cut-off frequency.
The current at points A and B =0.707 Im = IM/√2

Hence this two points A and B on the resonance curve are known as half power frequency points. Bandwidth may also be expressed as
BW = fr/Q
Where, fr = Resonant frequency
Q = Quality factor

PARALLEL RESONANCE IN AC CIRCUITS

Resistor (R), inductor (L) and capacitor (C) are connected in parallel across a sinusoidal voltage of variable frequency source. An R-L-C parallel resonance circuit is shown in the fig (a) and its vector diagram is shown in the fig (b).

IR = V/R, IL = V/XL, IC = V/XC

At resonance,
XL = XC
V/XL = V/XC
Therefore, IL = IC


Therefore, IL and IC are equal in magnitude but opposite in direction.

Resonant frequency

At resonance,

XL = XC

2 πfrL = 1 / 2 πfrC
Therefore, fr = 1 / 2 π√(LC)


PARALLEL RESONANCE (TWO BRANCH CIRCUIT)

A practical parallel circuit consisting of a coil of resistor (R) and inductor (L) is connected in parallel with a capacitor (C) as shown in the fig. (a). the vector diagram is shown in the fig (b).

The current in the coil (IL) is lagging behind the applied voltage V by an angle ФL.
The current in the capacitor (IC) leads the voltage by 90 .
For I is in phase with the applied voltage, the current IL sin ФL must be equal to IC.
At resonance, IC = IL sin ФL 
Resultant current at resonance

At resonance, the current in the circuit is in phase with the applied voltage.
From the vector diagram


L/CR is the equivalent impedance and is much larger than actual resistance. So parallel resonant circuit offers maximum impedance and the current is minimum.
Resonance curve

Impedance versus frequency curve

The impedance versus frequency curve of a two branch R.L.C parallel circuit is shown in the fig. the impedance of the circuit is maximum at resonance. As the frequency changes from resonance, the circuit impedance decreases rapidly.

For frequencies below resonance, the capacitive reactance XC is higher. Therefore, more current will flow through the coil. Thus the circuit behaves as inductive and the current lags behind the applied voltage.

For frequencies above resonance, XL is higher. Therefore, more current will flow through the capacitor. Thus the circuit behaves as capacitive and the current leads behind the applied voltage.

Current versus frequency curve

Fig shows the current versus frequency curve of the parallel circuit. The current at resonance is minimum. As the frequency changes from resonance, the circuit current increases rapidly.

Q-factor of a parallel resonant circuit

The ratio of circulating current between L and L to the supply current is known as current magnification or Q-factor of the parallel resonant circuit.

Comparison of series and parallel resonance

S. No.
Particulars
Series resonance circuit
Parallel resonance circuit
1
Resonant frequency
fr = 1/2 π√(LC)
fr = 1/2 π√[1/LC – (R/L)2]
2
Impedance at resonance
Minimum
Maximum
3
Current at resonance
Maximum
Minimum
4
Dynamic resistance
R
L/CR
5
Power factor at resonance
Unity (1)
Unity (1)
6
Q factor
XL/R or 1/R√(L/C)
XL/R or 1/R√(L/C)

 Applications of resonance:

1. It is used in tank circuit of an oscillator.
2. It is used in micro wave communication.
3. It is used in the tuning circuit of ratio and T.V to obtain the required station.


Sunday, 18 August 2019

Kirchhoff's Law with Solved Problems

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( a) Kirchhoff's First Law (or) current law (KCL)

The sum of currents flowing towards a junction would be equal to the sum of current flowing away from that junction.

(or)

The arithmetic sum of currents at a junction (node) of a network is zero.

Explanation

Assume five conductors carrying the currents I1, I2I3, I4 and I5 gathering at point ‘O’ as shown in figure. The arrow marks indicate the flowing direction of the current flowing through the conductor.

Assuming signs of currents flowing towards point ‘O’ as positive and the signs of currents flowing away from point ‘O’ as negative.

According to Kirchhoff's current law

(I1 ) + (- I2 ), + (I3 ) + ( I4 ) + ( -I5 ) = 0
I1 - I2 + I3 + I4 - I5 = 0
I1 + I3 + I4 = I2 + I5

Sum of incoming currents = Sum out going currents

( b) Kirchhoff's Second Law (or) voltage law (KVL)

In a closed circuit the algebraic sum of the potential drops would be equal to the algebraic sum of the potential rises.

(or)

The algebraic sum of voltage in a closed circuit (mesh) is equal to zero.

Explanation

(i) For e.m.f source ( battery)

(a) If we go through the positive point of the battery to the negative point, there is a fall (drop) in potential. Hence the sign would be negative.

(b) If we go towards the negative point of the battery to the positive point, there is a grow in potential. Hence the sign is positive.

(ii) For voltage drop (IR drop)

(a ) If we go towards the resistor in the same path as the current there is fall (drop) in potential ( because current flows from higher potential to lower potential) . Hence the sign is negative.

(b) If we go from the resistor opposite to the current direction , there is a rise in potential. Hence the sign is positive.

In figure ABCD is a closed loop. According to Kirchhoff's voltage law,

+ E - IR1 - IR2 = 0
(or)

E = IR1 + IR2

Therefore , E.m.f’s in the circuit = Voltage drop in the circuit

Solved Problems

Calculate the currents in different branches of the circuit shown in the figure

Solution
In the closed loop ABEFA ,

By using Kirchhoff's voltage law,
-3I1 - 2 ( I1 + I2 ) + 35 = 0
-3I1 - 2I1 - 2I2 + 35 = 0
-5I1 - 2I2 = - 35 ---------------- (1)

In the closed loop ABEFA,

By using Kirchhoff's voltage law,

-4I2 - 2 ( I1 + I2 ) + 40 = 0
-4I2 - 2I1 - 2I2 + 40 = 0
-2I1 - 6 I2 = - 40 ---------------- (2)

We can write the equations (1) and (2) in matrix form






















Current through 3 Ω resistor = I1 = 5 amps.
Current through 4 Ω resistor = I2 = 5 amps.
Current through 2 Ω resistor = I1 + I= 5+ 5 = 10 amps.