Module 2

Single-phase Circuits (Continued)

Three-phase Circuits

����������Single-phase Circuits (Continued)�

A.C.Through Series R-L Circuit

• Consider a circuit consisting of pure resistance R ohms connected in series with a pure inductance of L henries.
• The series combination is connected

across a.c. supply given by

v =V_m sin ωt

• Circuit draws a current I then there are two voltage drops,

a) Drop across pure resistance, V_R=IR

b) Drop across pure inductance, V_L =I X_L

where X_L=2πf L

I = r.m.s. value of current drawn

V_R,V_L= r.m.s. values of voltage drops

• The Kirchhoff’s law can be applied to the a.c. circuit but only the point to remember is the addition of voltages should be vector addition.

V = V_R + V_L

V = I R + I X_L

• The phasor diagram and the voltage triangle are shown as follows:

• In rectangular form the impedance is denoted as

Power

A.C. Through Series R-c Circuit

• Consider a circuit consisting of pure resistance R-ohms and connected in series with a pure capacitor of C-Farads.
• The series combination is connected across

a.c. supply given by v =V_m sin ωt.

• Circuit draws a current I, then there are

two voltage drops,

a. Drop across pure resistance, V_R = I R

b. Drop across pure inductance, V_C = 𝐼X_C

• The Kirchoff’s law can be applied to the a.c. circuit but

only the point to remember is the addition of voltages should be vector addition.

V = V_R + V_C

V = I _R + I X_C

The phasor diagram and the voltage triangle

Impedance

• Similar to R-L series circuit, in this case also, the impedance is nothing but opposition to the flow of alternating current.
• It is measured in ohms given by

A.C. Through Series R-L-C Circuit

• Consider a circuit consisting of resistance R Ohms, pure Inductance L Henries and capacitance C Farads connected in series with each other across a.c. supply. The circuit is as shown below.

• The a.c. supply is given by, 𝑣 = 𝑉_𝑚 𝑠𝑖𝑛𝜔𝑡
• The circuit draws a current I. Due to current I, there are different voltage drops across R, L and C which are given by,

a. Voltage across R, V_R = I R ... V_R is in phase with I.

b. Voltage across L, V_L = I X_L ... where V_L leads I by 90°.

c. Voltage across C, V_C = I X_C ... where V_C lags I by 90°

According to Kirchoff’s law we can write

V = V _𝑅 + V _𝐿 + V_C

• Case(i): 𝑿_𝑳 > 𝑿_C

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So, if 𝑣 = 𝑉_𝑚𝑠𝑖𝑛𝜔𝑡, then 𝑖 = 𝐼_𝑚𝑠𝑖𝑛(𝜔𝑡– 𝜙) as current lags voltage by angle 𝜙.

• Case(ii): 𝐗_𝐋 < X_C

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• The current is said to be capacitive in nature. The phasor sum of V_R and(V_C − V_L) gives the resultant supply voltage V. Current I leads the voltage V by 𝝓.

• Case (iii): 𝐗_𝐋=𝐗_𝐂
• When 𝐗_𝐋= 𝐗_𝐂, obviously 𝐕_𝐋= 𝐕_𝐂. So V_L and V_C will cancel each other and their resultant is zero.
• So, 𝐕_𝐑 = V in such case and overall circuit is purely resistive in nature.
• The phasor diagram for the above case is shown

�����Three-phase Circuits

Advantages of Three Phase System�

• The output of three phase machine is always greater than single phase machine of same size, approximately 1.5 times.
• For a given size and voltage a three phase alternator occupies less space and has less cost too than single phase having same rating.
• It is possible to produce rotating magnetic field with stationary coils by using three phase system.
• Three phase motors are self starting.

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Generation of Three Phase Voltage System�

• The ends of each coil are brought out through the slip

ring and brush arrangement to collect the induced e.m.f.

• Let e_R , e_Y and e_B be the three independent voltages in coil R₁R₂, Y₁Y₂ and B₁B₂ respectively.
• All of them will be displaced by one another by 120°.

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Star Connection�

• The star connection is formed by connecting starting or terminating ends of all the three windings together.
• The ends R₁ - Y₁ - B₁ are connected or ends R₂ - Y₂ - B₂ are connected together. This common point is called Neutral Point.

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Delta Connection�

• The delta is formed by connecting one end of winding to starting end of other and connections are continued to form a dosed loop.
• The supply terminals are taken out from the three junction points.

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Relation For Star Connected Load

• Line voltages V_L = V_RY= V_YB = V_BR
• While Line currents I_L = I_R= I_Y = I_B
• Phase voltages V_ph= V_R= V_Y=V_B
• Phase currents I_ph=I_R=I_Y=I_B

Also, From the circuit diagram

𝐈_𝐩𝐡= 𝐈_𝐋

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• The three phase voltages are displaced by 120° from each other.

The phasor diagram to get V_RY is shown.

• The V_Y is reversed to get −V_Y and then it is added to V_R to get V_RY.

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• The perpendicular is drawn from point A on vector OB representing V_L.

• In triangle OAB, the sides OA and AB are same as phase voltages. Hence OB bisects angle between V_R and −V_Y.

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Relation For Delta Connected Load

Measurement Of Power And Power Factor By Two Wattmeter Method

• Consider star connected load and two wattmeter connected as shown in figure.

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• Let us consider the RMS values of current and voltage to prove that sum of two wattmeter gives the total power consumed by the three phase load.
• W₁ = I_Rx V_RBx cos∠I_R^V_RB
• W₂ = I_Yx V_YBx cos∠I_Y^V_YB

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• Angle between V_Y and I_Y = φ
• V_ph= V_R= V_Y=V_B and V_RB= V_YB= V_L
• I_R= I_Y= I_L=I_PH(star)

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• From the vector diagram

Angle between V_RB and I_R = 30° - φ

Angle between V_YB and I_Y = 30° + φ

W₁ = I_RV_RBcos(30° - φ ) i.e. W₁ = I_LV_Lcos(30° - φ)

W₂ = I_YV_YBcos(30° + φ) i.e. W₂ = I_LV_Lcos(30° + φ)

W₁ + W₂ = I_L V_L [cos(30° - φ ) + cos(30° + φ) ]

= 2 I_L V_L (√3)/ 2 cos φ

𝐖_𝟏 + 𝐖_𝟐= √𝟑 𝐈_𝐋 𝐕_𝐋 cos φ = Total 3 phase power

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Power Factor Calculation By Two Wattmeter Method

In case of balanced load, the p.f. can be calculated from W₁and W₂ readings.

For balanced lagging p.f.

W₁ = I_LV_Lcos(30° - φ ),

W₂ = I_LV_Lcos(30° + φ)

𝐖_𝟏 + 𝐖_𝟐= √𝟑 𝐈_𝐋 𝐕_𝐋 cos φ (i)

𝐖_𝟏 − 𝐖_𝟐= 𝐈_𝐋 𝐕_𝐋 sin φ (ii)

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• Taking ratio of equation (ii) and (i)

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Effect Of Power Factor On Wattmeter Readings

• For balanced lagging p.f.

W₁ = I_L V_Lcos(30° - φ ), W₂ = I_LV_Lcos(30° + φ)

• Consider different cases

Case (i) cosφ = 0 i.e. φ=90°

W₁ = I_L V_Lcos(30° - 90°) = ½ I_L V_L

W₂ = I_LV_Lcos(30° + 90°) = - ½ I_L V_L

i. e. 𝐖𝟏 + 𝐖𝟐 = 0

|𝐖𝟏| = |𝐖𝟐| but 𝐖𝟐 = - 𝐖1

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• Case (ii) cos φ = 0.5 i.e. φ= 60°

W₁ = I_LV_Lcos(30° - 60°) = √3/2 I_L V_L

W₂ = I_LV_Lcos(30° + 60°) = 0

𝐖_𝟏 + 𝐖_𝟐 = 𝐖_𝟏 = Total Power

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• Case (iii) cosφ =1 i.e φ = 0

W₁ = I_L V_Lcos(30° - 0°) = √3/2 I_L V_L

W₂ = I_LV_Lcos(30° + 0°) = √3/2 I_L V_L

Both wattmeters read equal and positive

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