Q1) Three similar resistors are connected in star across 400V 3-phase lines. Line current is 4A. Calculate the value of each resistor.

A1) For star connection:

IL=Iph=4A

Vph=VL/ = 400/ = 231V

Rph= 231/4= 57.75ohm

For Delta Connection:

IL=4A

Iph= IL/

=4/ ==2.30A

Zph=400/2.30=173.9ohm

Rph= 173.9/3 = 57.97ohm

Q2) Three identical impedances are connected in delta 3-phase supply of 400V. The line current is 30A and total power taken from the supply is 10kW. Calculate the resistance and reactance value of each impedance?

A2) VL=Vph=400V

IL=30A

Iph=IL/= 30/ =17.32A

Zph=Vph/Iph= 400/17.32=23.09ohm

P=VLIL Cos Ø

Cos Ø = 10000/ 400x30 = 0.48

Sin Ø =0.88

Rph=Zph Cos Ø= 23.09x0.48=11.08ohm

Xph=Zph Sin Ø = 23.09x0.88=20.32ohm

Q3) A star connected alternator supplies a delta connected load. The impedance of the load branch is 6+j5 ohm/phase. The line voltage is 230V. Determine the current in the load branch and power consumed by the load.

A3) Zph= = 7.8ohm

VL=Vph=230V

Iph=Vph/Zph=230/7.8=29.49A

Iph=IL/

IL= Iph=x29.49=51.07A

P=VLIL Cos Ø = x 230x51.07x0.768=15.62kW

Q4) The load connected to a 3-phase supply comprise three similar coils connected in star. The line currents are 25A and the kVA and kW inputs are 18 and 10 respectively. Find the line and phase voltage, the kVAR input resistance and reactance of each coil?

A4) IL= 25A

P= 10000W

Cos Ø = 10/18 = 0.56

P=VLIL Cos Ø

10000= x VLx25x0.56

VL =412.39V

Vph= VL/ = 412.39/=238.09V

KVAR= = 14.96

Zph=238.09/25=9.52ohm

Rph=Zph Cos Ø= 9.52x0.56=5.33ohm

Xph=Zph Sin Ø = 9.52x0.83=7.88ohm

Q5) A balanced delta connected load consisting of three coils draws 8 A at 0.5 p.f from 100V 3-phase ac supply. If the coils are reconnected in star across the same supply. Find the line current and total power consumed?

A5) For Delta connection:

IL=8A

Iph= IL/= 8A

Vph=100V

Zph=100/8=12.5ohm

Rph=Zph Cos Ø=12.5x0.5 = 6.25ohm

Xph=Zph Sin Ø = 12.5x0.866=10.825ohm

P=VLIL Cos Ø

= x 100x 8x0.5=1200W

For Star Connection:

Vph= VL/= 100/V=57.73V

Zph=100/8=12.5ohm

Iph=57.73/12.5=4.62A

P=VLIL Cos Ø

= x 100x 4.62x0.5

P= 400W

Q6) A 3-phase supply with a line voltage of 250V has an unbalanced delta connected load as shown in figure

Find (a) Phase current (b) line current?

A6) From above figure we can write

The phase currents are

The Line currents are

Q7) Find vc(t) for circuit shown in the figure?

A7) Let the current in circuit be i(t). From KVL we have

Taking F.T we get

As i(0+) =0 hence I(0)=0

Takin IFT we get

Q8) Find Fourier transform of the waveform given below?

A8)

Q9) Find the Fourier series coefficients for the given saw tooth wave(in exponential form) as shown in figure

A9)

Q10) A continues time periodic signal x(t) has a fundamental period T=8. The nonzero Fourier series coefficients for x(t) are: 𝑎1 = 𝑎−1 = 2. 𝑎3 = 𝑎∗−3 = j4

Express x(t) in the form of

A10)

Here,

Therefore,

Q11) For the circuit shown in figure, Determine the voltage response 𝑣𝑜 𝑡 to a current source excitation i(t)=2𝑒−𝑡𝑢(𝑡), using Fourier transform?

A11) We apply KCL

Taking FT of both sides we have

Therefore,

By partial fraction of above equation, we get

Taking IFT of above equation we get

Q12) In the Circuit of fig., A 400 V, 50 Hz, three phase supply of phase sequence ABC is supplied to a delta connected load consisting of a 100 Ohm resistor between lines A and B, a 378 mH inductor between lines B and C, and a 37.8 micro farad capacitor between lines C and A. Determine Phase and line currents.

A12) VL = 400V, f=50Hz, R= 100 Ohm, L= 318mH, C=31.8 micro farad

Phase Current

Line Currents

Q13) For a series RLC circuit having R=10ohms, L= 0.15H, C=100F. They are connected across 100v 50Hz supply. Calculate total impedance?

A13) Impedance Z=

Z= = 18.27ohm

Q14) For a series RLC circuit having R=12ohms, L= 0.2H, C=60F. They are connected across 100v 50Hz supply. Calculate circuit current?

A14) I=

Z=

Z= = 13.89ohm

I = 100/13.89 =7.2A

Q15) For a series RLC circuit having R=10ohms, L= 0.15H, C=100F. They are connected across 100v 50Hz supply. Calculate power factor?

A15) cosφ =

Impedance Z=

 Z= = 18.27ohm

Cosφ = =

φ = 56.81o lagging

Q16) If V1=6cos2t, find i2 at steady state.

A16) Apply KVL

6-sI1(s)-2I2(s)=0

-2I2(s)-I2(s)+ I1(s)=0

I2(s)= I1(s)

I1(s)=[I2(s)

Substituting I1(s) in first equation

6-s[}I2(s)- 2I2(s)=0

Solving above equation we get

I2(s)=

Replace s=jω, s=2j

I2(jω)===

=

Q17) If V1=2sin2t, Va at steady state will be?

A17) Zeq=

By voltage divider rule Va=

Replace s by jω , s=2j

Va==

Va==

Va=

Q18) Find the Fourier integral of

f(x) = |sin x|               |x| ≤ π

= 0                          |x| ≥ π 

Deduce that π +1/ 1 - 2  cos (π/2) d = π/2

A18)

f(x) = 2/π 

=t   =

=

= - cost(1-) ]0 π  - cost(1+)   ]0 π

2(1-)                 2(1+)

= 1/ 1-  2 [cos π+ 1]

=π +1/ 1 - 2  cos (π/2) d = π/2

Q19) Find the Fourier Integral of

f(x) = 1        |x| ≤ 1

A19)

f(x) = 1/π   (t-x) dt d

= 1/π  dt d

= 1/π /   ] -1 1     d

= 1/π  - /  d 

= 1/π – sin ]/  d 

= 2/π   /  d       = π/2  when |x| < 1 and 0 when |x| >1

By setting x=0

= /    d  = π/2.

Q20) Compute the Exponential series of the following series.

A20)

The time period of the signal x(t) is T=4.

Ω0 = 2π/T = 2 π/4= π/2

C0 = 1/T = ¼ [ +

C0 = ¼ [ 2+1] = ¾

Cn = 1/T [    e-jnΩot dt = ¼[  e-jnπ/2t dt + e-jn π/2t dt

= ½. 1/ -jn π/2 [e-jn π/2t ] 0 1 + ¼ 1/-jn π/2[ e-jn π/2t ]  1 2

= - 1/jn π[e-jn π/2  – 1] – 1/ 2jn π (e-jn π – e-jn π/2 )

= 1/jn π[ 1 - e-jn π/2 ] – 1/ 2 e-jn π  + ½ e-jn π/2 )

= 1/jn π [ 1- ½ (-1) n -1/2 e-j n π/2 ]