December 2019 Solutions

Q.1. Solve the following :- [20 M]

Ans.

Given :-

To Find :- I=?

Solution :-

Step-I :- Using source transformation and converting the 10 V voltage source into current source, so now it becomes

I_{s}=\frac{V}{R}=\frac{10}{2}= 5 \; \; A

So by converting the voltage source in current source we will get two current sources . one is 5 A and second is 6 A.

Step-2:- Now using current divider rule and calculating the value of I

I= \frac{\frac{1}{4}}{\frac{1}{2}+\frac{1}{6}+\frac{1}{4}}

I=\frac{1/4}{0.91}= \frac{0.25}{0.91}=0.274 A.

Step-3 :- As there are two current source so we will add the two current source

so it becomes I_st = I_s1 + I_s2 = 5 + 6 = 11 A

and now the circuit becomes

Step-4 :- Finally calculating the current through 4 Ω is

I= 0.274 x 4 = 1.096 A

I= 1.096 A

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Ans.

Let N₁= No. of turns in primary

N₂ = No. of turns in secondary

Φ_m = maximum flux in core in webers = B_m x A

f = frequency of a.c input in Hz

Flux increases from its zero value to maximum value Φ_m in one quarter of cycle . i.e. in 1/4 second

Therefore, Average rate of change of flux = \frac{\phi {m}}{1/4 f }= 4 f \phi {m} \; Wb/s \; or \; volt

Now, rate of change of flux per turn means induced emf in volts .

Average e.m.f/turn = 4 f Φ_m volt

If flux Φ varies sinusoidally, them r.m.s value of induced e.m.f is obtained by multiplying average value with form factor .

Form factor = \frac{r.m.s \; value }{average \; value }= 1.11

r.m.s value of e.m.f turn = 1.11 x 4 f Φ_m = 4.44 f Φ_m volt

Now, r.m.s value of induced e.m.f in whole of primary winding = (induced e.m.f/turn) x No. of Primary

E₁= 4.44 f N₁ Φ_m = 4.44 f N₁ B_m A turn

E₁ = 4.44 f N₁ B_mA ——(1)

Similarly, r.m.s value of induced e.m.f in secondary is

E₂ = 4.44 f N₂ B_mA ——–(2)

From (i) and (ii)

\frac{E_{1}}{N_{1}}=\frac{E_{2}}{N_{2}}= 4.44\; f\; \phi _{m}

This means that e.m.f/turn is same in both primary and secondary windings.

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Ans.

Consider a three phase alternator with three armature coils a,b,c which is displaced by 120^o apart from one another.

The three coils have three emfs induced in them which are similar in all respects and they are 120^o out of phase with one another. Each voltage wave is assumed to be sinusoidal and have maximum value of E_m.

As three circuits are similar but are 120^o electrical degrees apart, the emf waves generated in them are displaced from each other by 120^o .

As the waves are sinusoidal , so the emf in phase ‘a’ is zero at any time instant. So, the instantaneous values of three emfs will be given as

e_a= E_m sin ωt ——-(1)

e_b= E_m sin (ωt-120^o) —–(2)

e_c= E_m sin (ωt-240^o) ——(3)

The actual values of these voltages vary from peak positive to zero and to peak negative values in one revolution of the vectors.

The sum of the above three equations (1), (2) and (3) is zero and i can be shown as

Resultant instantaneous emf = e_a + e_b + e_c

= E_m sin ωt + E_m sin (ωt-120^o)+E_m sin (ωt-240^o)

= E_m [sin ωt + 2 sin (ωt-180^o)cos 60^o)]

= E_m [ sin ωt-2 sin ωt cos 60^o] = 0

Hence, it is proved that the sum of three phase emf’s is zero in a three phase ac circuit.

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Ans.

Sr. No.	Series resonant circuit	Parallel resonant circuit
1.	At the resonant frequency, the resistance, inductance and capacitor connected in series contains the minimum impedance .	At the resonant frequency, the resistance, inductor and capacitor connected in parallel contains maximum impedance .
2.	The series resonant circuit is called as the accepter circuit.	The parallel resonant circuit is called as the rejector circuit.
3.	In series resonant circuit, the impedance is minimum and is equal to the resistance in the circuit. i.e. Z=R	In parallel resonant circuit, the impedance is maximum and nearly equal to infinity. i.e. Z= \infty .
4.	In the series resonant circuit, the current is maximum at the resonance.	In the parallel resonant circuit, the current is minimum at the resonance.
5.	The series resonant circuit magnifies the voltage.	The parallel resonance circuit magnifies the current.
6.	In series circuit power factor is unity.	In parallel resonance power factor is unity.
7.	In series circuit the resonant frequency is given by 1/2\pi \sqrt{LC} .	In parallel circuit the resonant frequency is given by \frac{1}{2\pi }\sqrt{\frac{1}{LC}-\frac{R^{2}}{L^{2}}} .

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Ans.

Given :- Coil A

I_A= 2 A, cos Φ = 0.8 , V_A= 10 V

Coil B

I_B= 2 A , cos Φ = 0.7 (lag) , V_B= 5 V

To Find :- Voltage across series circuits= ? , P.f = ?

Solution :-

i) Current in coil A, I_A= I_A cos φ- j I_A sin φ

= 2 x 0.8 – j2 x 0.6

= 1.6 – j 1.2 A

Impedance of coil A,

Z_{A}=\frac{V+j0}{I_{A}}=\frac{10+j0}{1.6-1.2j }= (4+3j)\Omega

Similarly, The current in coil B, I_B= I_B cos φ – jI_B sin φ

= 2 x 0.7 – j2 x 0.714

= 1.4- j 1.428 A

Voltage applied to the coil B, V_B= 5 V

Impedance of the coil B = Z_{B}=\frac{V+j0}{I_{B}}=\frac{5+j0}{1.4-1.428j }= (1.75+1.77i)\Omega

a) When the two coil are in series, the total impedance of the series circuit is given by

Z= Z_A+Z_B (phasor sum)

Z= (4+3j) + ( 1.75-1.77j)

Z= 5.75 + j 4.77

Z=\sqrt{(5.75)^{2}+(4.77)^{2}}= 7.5\; \Omega

Z= 7.47 Ω

The current given in the series circuit is given = 2 A

Voltage required across the series circuit = IZ

V= 2 x 7.47 = 14.94 ≈ 15 V

Voltage across the series circuit =15 V

b) Power factor in the series circuit

For finding power factor we need to find the total current drawn by the series circuit

I = I_A + I_B

= (1.6 – j 1.2 ) + 1.4- j 1.428

= (1.6 + 1.4 ) -j( 1.2 +1.428)

= 3-j(2.628)

I = I_active – j I_reactive

So here , I_active = 3 and I_reactive = 2.628

Hence, angle of lead or lag of this current is given by

\phi =tan^{-1}\frac{I_{reactive}}{I_{active}} tan \; \phi =\frac{I_{reactive}}{I_{active}}

\phi =tan^{-1}\left (\frac{I_{reactive}}{I_{active}} \right )= tan^{-1}\left ( \frac{2.628}{3} \right )=41.21^o

Power factor of the circuit as a whole , cos φ = cos 41.21^o = 0.752(lagging)

cos φ = 0.752

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Ans.

Phasor diagram –

i) The phasor diagram of a transformer on no-load is shown in the figure. Here Io current is divided into two components Io sin φ_o and cos φ_o. The first component i.e. I_o sin φ_o is in the direction of flux and hence it maintains the flux in the core. This component is called the magnetising component (I_μ)

I_μ= I_o sin φ_o

ii) The second component Io cos φ_o is in phase with V1 and it supplies the small amount of cu loss and the iron loss. This is called a sthe working component or iron loss component (I_w) .

I_w= I_o cos φ_o

The no load component is gicven by,

I_{o}=\sqrt{I_{w}^{2}+I_{\mu }^{2}}

The function of first component is to sustain the alternating flux in the core of the transformer, which is responsible for the emf induced in secondary winding of the trasnformer. The magnetising current is wattless component. As Io is very small, the primary Cu loss can be neglected and hence all the input power at NO load is equal to the iron loss or core loss of trasnformer.

Iron loss = V₁ I_o cos φ_o watts.

and p.f = cos φ_o (lag)

Io always lag behind V₁ by φ^o.

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Ans.

Working principle of DC motor :- The workling of DC motor is based on the principle that Whenever a current carrying conductor is placed in a magnetic field it experiences a force. An electric motor is a machine that converts an electric energy into mechanical energy.

Working principle of DC generator :- An electrical generator is a machine which converts mechanical energy into electrical energy. The energy conversion depends on the principle of production of dynamically induced emf . Whenever a conductor cuts magnetic flux, dynamically induced emf is produced in it according to Faraday’s Law of Electromagnetic Induction. This emf causes a current to flow if the conductor circuit is closed. So, in simple we can say that whenever a conductor cuts magnetic lines of force, an emf is induced in the conductor.

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Ans.

Let us open the 20 Ω ressitance

Let us find thevenin’s equivalent across AB.

Now finding Rth seen through AB by replacing sources by their internal resistances.

R_TH = 6.67 Ω

Now, let us find V_th across A and B .

Take 3 loops as shown in the figure,

Writing KVL for combined loop 1 and 2.

12-2I₁ – 10I₁ – 5I₂ – 15(I₂-I₃) = 0

Simplifying we have,

-12 I₁– 20 I₂ + 15 I₃ = -12

Writing KVL for loop 3

-25 I₃ – 8 – 5I₃ – 15(I₃-I₂) = 0

Simplifying we have ,

0 I₁ + 15 I₂ – 45 I₃ = 8 —(2)

From the 4 A branch

4 = I₁– I₂

I₁– I₂ + 0 I₃ = 4 —(3)

Solving (1),(2) and (3)

I₁= 2.56 A

I₂ = -1.43 A

I₃ = -0.65 A

Now, consider the part of network,

V_AB = -10 I₁ – 2I₁ + 12

= 12 I₁ +12

= 12 x 2.56 +2 = -18.72 V (B positive w.r.t A)

V_th = 18.72 V , (B positive w.r.t A)

Drawing Thevenin’s equivalent across AB

Now, consider 20 Ω across AB .

I= \frac{18.72}{6.67+20}=0.701 A

V₂₀ = 0.701 x 20 = 14.03 V

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Ans.

Given :- W₁/W₂= 2/1 i.e. W₁= 2W₂

To Find :- Power factor of the system

Solution :-

\phi =tan^{-1}\left [ \frac{\sqrt{3}(W_{1}-W_{2})}{W_{1}+W_{2}} \right ]= tan^{-1}\frac{\sqrt{3}W_{2}}{3W_{2}}

φ= tan^-1(0.557) = 30^o

Power factor = cos φ = cos 30^o = 0.8661

So, the power factor of the system is 0.8661.

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Ans.

Given :-

Period of wave , T=4 sec

To Find :- RMS value of the waveform

Solution :-

I_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}i(t)^{2}dt} i(t)= \left\{\begin{matrix}20/2 & t=10t & 0<t<2\\0 & & 2<t<4\end{matrix}\right. I_{rms}=\sqrt{\frac{1}{4}\left [ \int_{0}^{2} (10 t)^{2}dt+\int_{2}^{4}0^{2}\; dt\right ]} I_{rms}=\frac{1}{2}\sqrt{\int_{0}^{2}100t^{2}dt} I_{rms}=\frac{10}{2}\sqrt{\left (\frac{t^{3}}{3} \right )^{2}_{0}} I_{rms}=\frac{5}{\sqrt{3}}\sqrt{2^{3}-0^{3}}=\frac{5}{\sqrt{3}}\sqrt{8}= 8.164\; A

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Ans.

Given :- C= 2.5 μF , R= 15 Ω , L= 260 mH

To Find :- i) Resonant frequency f_o

ii) Q factor of the circuit at resonance

iii) dynamic impedance of the circuit

Solution :-

i) To find the resonant frequency of the parallel circuit , the formula is given as

f_{o}=\frac{1}{2\pi }\sqrt{\left ( \frac{1}{LC}-\frac{R^{2}}{L^{2}} \right )} f_{o}=\frac{1}{2\pi }\sqrt{\left ( \frac{1}{260\times 10^{-3}\times 2.5\times 10^{-6}}-\frac{15^{2}}{(260\times 10^{-3})^{2}} \right )}

f_o= 9.17 Hz

ii) To find the Q factor

Q=\frac{\omega L}{R}=\frac{2\pi f_oL}{R}= \frac{2\pi \times 9.17\times 260\times 10^{-3}}{15}= 0.99\approx 1

Q=1

iii) The dynamic impedance of the circuit

Y=1R+ jωC – jωL

Y= 1 x 15 + j x 2π x 9.17 x (2.5 x 10^-6) – j x 2π x (260 x 10^-3)

Y = 15 + j(1.44 x 10^-4 ) – j (1.498)

Y= 15-1.4978 j

Y=15.074\angle -5.702

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Ans.

Given :- Z= (14.151 – j200) Ω , V= 400 V, f = 50 Hz

To Find :- i) branch current

ii)n line current

iii) Total power taken

Solution :-

i) Finding branch current , line current , phase voltage and line voltage

Line voltage be V_L= 400 V

Let phase voltage be V_AB, V_BC, V_CA

V_AB= 400 ∠0^o

V_BC= 400∠-120^o

V_CA= 400 ∠120^o

Let phase current be I_A, I_B, I_C

Phase current is given by

I_{A}=\frac{V_{AB}}{Z}=\frac{400\angle 0}{(14.151-j200))}=0.1408 + 1.99j= 1.99\angle 85.95 I_{B}=\frac{V_{AB}}{Z}=\frac{400\angle -120}{(14.151-j200))}=1.653-1.116j= 1.99\angle -34.04

I_{C}=\frac{V_{CA}}{Z}=\frac{400\angle 120}{(14.151-j200))}=-1.7938-0.8730j = 1.99\angle -154.04 A

As it is a delta connection we know that

I_{L}=\sqrt{3}I_{ph} —-(1)

I_{ph}=\frac{V_{ph}}{Z_{ph}}=\frac{400}{200.50}=1.99\approx 2 A

Putting value of I_ph in eq.(1)

I_{L}=\sqrt{3}I_{ph}= \sqrt{3}\times 2= 3.46 A

Finding Line current

I₁= I_A-I_B

= (3.45 ∠ 85.95)-(1.99∠-34.04) = -1.4053+4.55j = 4.76 ∠ 107.14^o A

I₁= 4.76 ∠ 107.14^o A

I₂=I_B-I_C

I₂= (1.99∠-34.04)- (3.45 ∠ 85.95) = 1.4053-4.55j = 4.76∠ -72.85 A

I₂= 4.76∠ -72.85 A

I₃= I_C– I_A

= (1.99∠-154.04)-(3.45 ∠ 85.95) = -2.0328+4.31j = 4.76 ∠ -115.23^o A

I₃= 4.76 ∠ -115.23^o

ii) Finding total power taken

For that we need Power factor

P.F= cos φ = R/Z

Z=\sqrt{(14.151^{2}+(200)^{2}}=200.50

P.F= cos φ = R/Z = \frac{14.151}{200.50}=0.070

P= 3 x V_ph x I_ph x cos φ

P= 3 x 400 x 2 x 0.070

P= 168 W

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Ans.

The given diagram is

As C and F are connected by wire. Let us rotate circuit by 90 degrees anticlockwise.

R_AB = 10.5 Ω

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Ans..

Given :-

To Find :- Current through 6 Ω

Solution :-

Step-1 :-Transforming 240 V and 60 V voltage source into current source. Marking node O, A , B ,C on diagram . Self and mutual conductance terms can be written down .

Converting the 60 V and 240 V voltage source into current source

V₁=IR

60 = I x 12

I₁= 5 A

V₂= IR

240 = I x 3

I₂= 80 A

At A, G_aa= 1/3 + 1/6 = 0.5

At B, G_bb= 1/30 + 1/6 = 0.2

At C, G_cc = 1/12 = 0.0833

Between A and B, G_ab = 1/6 = 0.16

Between B and C , G_bc = 1/12 = 0.0833

Between A and C ,G_ac = 0.33

Current source matrix

At node A, 80 A is incoming and 10 is outgoing currents give a net outgoing current of 70 A

At node C, the incoming current is 5 A . At node B there is no current source connected. Hence the current matrix is given as

\begin{bmatrix}70\\0\\10\end{bmatrix}

THe voltages af three nodes are found to be V_a, V_b, V_c

Now, the matrix becomes

\begin{bmatrix}0.5 & -0.16 & 0 \\-0.16 & 0.2 & -0.083\\0 & -0.083 & 0.083\end{bmatrix}\begin{bmatrix}V_{a}\\V_{b}\\V_{c} \end{bmatrix}=\begin{bmatrix} 70\\0\\10\end{bmatrix}

Solving the matrix we get

Δ= 0.5(0.0166-0.00688)+0.16(-0.01328-0)+0(-0.01328-0)

Δ=0.5(0.0098)+0.16(-0.01328)

Δ=0.0049-0.00212

Δ = 0.00278

ΔV_a = \begin{bmatrix}70 & -0.16 & 0 \\0 & 0.2 & -0.083\\10 & -0.083 & 0.083\end{bmatrix}

ΔV_a = 70(0.0166-0.0068)+0.16(0-0.83)+0(0+0.83)

= 70(0.0098)+0.16(-0.83)

= 0.688-0.1328

ΔV_a = 0.5552

ΔV_b= \begin{bmatrix}0.5 & 70 & 0 \\-0.16 & 0 & -0.083\\0 & 10 & 0.083 \end{bmatrix}

ΔV_b= 0.5[0-(-0.83)]-70(-0.01328-0)+0(-0.01328-0)

= 0.5(0.83)-70(-0.01328)

ΔV_b=1.3446

ΔV_c= \begin{bmatrix}0.5 & -0.16 & 70 \\-0.16 & 0.2 & 0\\0 & -0.083& 10\end{bmatrix}

ΔV_c= 0.5(2)+0.16(-1.6)70(0.01328)

=1-0.256+0.9296

ΔV_c=1.6736

So, V_a= ΔV_a/Δ = 0.5552/0.00278 = 199.71

V_b= ΔV_b/Δ = 1.3446/0.00278 = 483.66

V_c= ΔV_c/Δ = 1.6736/0.00278 = 602.01

Voltage across 6-ohm resistor is = V_b-V_a = 483.66-199.71 = 283.95 V

Current through 6 ohm resistor is

I= Voltage across 6 ohm resistor /6 = 283.95/6 = 47.32 A

Hencen current through 6 ohm resistance is 47.32 A

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Ans.

Given :-

To Find :- I₁ and I₂

Solution :-

Z₁= R+jX_L1 = 10 +j8 = 12.80∠38.65^o

Z₂= R₂ + jX_C = 9-j6 = 10.81∠-33.69

Z₃ = R+jX_L = 3+j2 =3.60∠33.69

The voltage across the parallel circuit will be considered as the V_c=100∠0

Vc=I₃/Z₃

100∠0 = I₃/3+2j

I_c=V_c/Z_c = 100∠0/3+2j = 23.07-15.38j = 27.73∠-33.69^o A

Now finding the Z_AB

Z_AB= Z₁ || Z₂ = 10 +j8 || 9-j6 = \frac{(10+j8)\times (9-j6)}{(10+j8)+(9-j6)}=138+12j = 138.52\angle 4.96^{o}

V_AB=I_C x Z_AB = (23.07-15.38j) x (138+12j) = 0.156-0.128j = 0.201∠-39.37^o

V_AB= 0.201∠-39.37^o

Now calculating the total voltage

V= V_c+V_ab = 100∠0+0.201∠-39.37^o = 100∠-0.072 V

V= 100∠-0.072 V

Now finding the current I₁ and I₂ , we will take voltage V_AB as they both are connected between points A and B

I₁=V_AB/Z₁ = 0.201∠-39.37^o /12.80∠38.65^o= 0.0157∠-78.02^o A

I₂ = V_AB/Z₂ = 0.201∠-39.37^o / 10.81∠-33.69 = 0.0185 ∠-5.68^o A

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Ans.

Given :- V₁= 440 V , V₂ = 220 V , KVA = 10 KVA , f=50 Hz , R_p=R₁= 0.2 Ω , X_p=X₁= 0.6 Ω , R_s=R₂= 0.04 Ω , X_s= X₂= 0.014 Ω, cos φ= 0.8

φ = cos^-1 (0.8) = 36.86 , Sin φ= 0.6

To Find :- Regulation on full load = ?

Solution :- K= V₂/V₁ = 220/440 = 0.5

KVA = 10 KVA,

R₀₂=R₂+K²R₁ = 0.04 + (0.5)² x 0.2 = 0.09 Ω

X₀₂ = X₂ + K²X₁ = 0.014 + (0.5)² x 0.6 = 0.164 Ω

I_{2}=\frac{V_{2}}{R_{2}}= \frac{220}{0.04}=5500 \; A

Voltage drop = I₂(R₀₂ cos φ ± X₀₂ Sin φ) = 5500 ( 0.09 x 0.8 ± 0.164 x 0.6) = 5500(0.072 + 0.0984) = 5500 x 0.1704 = 937.2 volts

As the power factor is laaging , so there will be positive sign . So, the voltage regulation can be calculated as ,

Voltage\; Regulation = \frac{I_{2}(R_{02}\; cos\phi + \; X_{02}\; sin\phi )}{V_{2}}\times 100

% Voltage Regulation = \frac{937.2}{220}\times 100 = 426\; %

% V.R = 426 %

The regulation on full load for 0.8 lagging power factor is 426 %.

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Ans.

Consider the three equal impedances (load) connected in star across star connected supply . As all three loads are equal therefore it is called as balanced load.

Assume phase sequence as RYB.

So, the voltage across three phases is given by

V_{R}=V\angle 0 V_{Y}=V\angle -120 V_{B}=V\angle -240

There is a 120^o phase difference between three voltages. The voltage induced in each winding i.e RYB is called as phase voltage and the current in each winding is called phase current . WHEREAS the voltage induced between any pair of terminals is called line voltage and the current flowing in each line is called line current.

Relation between line current and phase current

The line current is given by

I_{R}=\frac{V_{R}}{Z}=\frac{V\angle 0}{|Z|\angle \phi }= \frac{V}{|Z|}\angle -\phi I_{Y}=\frac{V_{Y}}{Z}=\frac{V\angle -120}{|Z|\angle -\phi }= \frac{V}{|Z|}\angle -120-\phi I_{B}=\frac{V_{B}}{Z}=\frac{V\angle -240}{|Z|\angle \phi }= \frac{V}{|Z|}\angle -240-\phi

All the three current have same magnitude and have 120^o phase difference . Current through neutral wire is given by

I_N=I_R+I_Y+IB —— (1)

Putting values in eq. (1)

I_{N}= \frac{V}{|Z|}\angle -\phi + \frac{V}{|Z|}\angle -120-\phi + \frac{V}{|Z|}\angle-240 -\phi

I_N=0

As the sum of three vectors have same magnitude but 120o phase difference it is always zero. Hence, current through neutral is also zero.

In star connection, line current is equal to phase current

I_L=I_Ph

Relation between line voltage and phase voltage

Let V_RY be the voltage between R and Y

V_RY= -V_Y+ V_R

We obtain V_RY by adding -V_Y and V_R

The resultant V_RY will be 30^o ahead of VR. Hence, V_L is 30o ahead of V_ph. The manitude of V_RY can be calculated as,

| V_{RY}|=\sqrt{|V_{R}|^{2}-|V_{Y}|^{2}+2|V_{R}|.\; |V_{Y}|\; cos \; 60\; }

AS we know, |V_R|= |V_Y| = |V_B| = V volts

| V_{RY}|=\sqrt{V^{2}+V^{2}+2V.V. \; cos 60} | V_{RY}|=\sqrt{3V^{2}}

|V_RY|= \sqrt{3}\; V=\sqrt{3}\; V

V_{L}=\sqrt{3}\; V_{ph}

Hence, the relation can be derived as the line voltage is \sqrt{3} times the phase voltage.

Now calculating the total power consumed

For 3-phase load

P= 3 x phase power

P=3\; V_{ph}\; I_{ph}\; cos\; \phi

But, V_{ph}=\frac{V_{L}}{\sqrt{3}}

and I_L=I_ph

P=\sqrt{3}\; V_{L}\; I_{L}\; cos\; \phi

This is the expression of power when it is a star connection.

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Ans.

Given :-

To Find :- V=?

Solution :-

Step-I :- Applying KVL to loop 1 , Consider current I₁ flowing in loop 1

20-6I₁+2(I₁-I₂) -5(I₁-I₃) -V=0

20-6I₁+2I₁-2I₂-5I₁+5I₃ – V= 0

-9I₁ -2I₂+5I₃ = 20+V —–(1)

Step-2 :- Applying KVL to loop 2, Consider current I₂ flowing in loop 2

-2(I₂-I₁) -3I₂-1(I₂-I₃) = 0

-2I₂ +2I₁ – 3I₂ – I₂+I₃ = 0

2I₁ -6I₂ + I₃ = 0 —(2)

Step-3 :- Applying KVL to loop 3 , Consider current I₃ flowing in loop 3

V-5(I3-I1) +1(I₃-I₂) -4I3 = 0

V- 5I₃ + 5I₁ + I₃-I₂ – 4I₃ = 0

V-8I₃+5I₁-I₂ = 0

5I₁-I₂-8I₃ = V —-(3)

Solving (1) ,(2) and (3) by using cramer’s rule

\begin{vmatrix}-9 & -2 & 5\\ 2 & -6 & 1\\ 5 & -1& -8\end{vmatrix}\begin{vmatrix}I_{1}\\ I_{2}\\ I_{3}\end{vmatrix}=\begin{vmatrix}20+V\\0\\V\end{vmatrix}

By solving the determinant we get

Δ=-343

Now to find V calculating ΔI₁

\Delta I_{1}=\begin{vmatrix}20+V & -2 &5 \\0 & -6 & 1\\V & -1 & -8\end{vmatrix}

Solving this we get

V= 61.25

In this way , the value of V is 61.25 V.

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Ans.

Given :- v= 100 sin 3t , C= 1/6 F , L= 1/3 H , R₁= 1 , R₂= 1

To Find :- branch current I₁ and I₂

Solution :-

As v= V_m sin ωt

Now, comparing with

V= Vm sin (ωt ± φ)

V_m=100

ω=3

i) For Branch I

Z₁ = R + jX_L = R+jωL

= 1+ j 3 x \frac{1}{3} Ω

= 1+ j1 Ω = \sqrt2 \angle 45^o Ω

I_{1}=\frac{V}{Z_{1}}=\frac{100\angle 0^{o}}{\sqrt{2}\angle 45}=70.71 \angle -45^{o}\; \Omega

ii) For Branch 2 ,

R=1 Ω , C= \frac{1}{6}

Z₂=R-jXc = R-j\frac{1}{\omega C}

Z_{2}=1-j\times \frac{1}{3\times \frac{1}{6}}=1-j2\Omega =\sqrt{5}\angle -63.43^{o}\Omega

I_{2}=\frac{V}{Z_{2}}=\frac{100\angle 0}{\sqrt{5}\angle -63.43^{o}}=44.72\angle 63.43^{o} A

I_{2}=44.72 \; Sin (3t+63.43^{o})A

Total current I,

I= I₁+ I₂

= 70.71 \angle -45^{o} + 44.72\angle 63.43^{o} A

= 70.71 \angle -8.13^o

I=70.71 \; Sin (3t-8.13^{o})A

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