2024, ഡിസംബർ 4

EE-3033-FEC-CO3.2 Revision Test -2

Coaching Based Revision-CBR-CO3.1- Revision Test -1

EE 3033 FEC CO4 M4 03 STAR Q No 7

EE 3033 FEC CO4 M4 03 STAR Q No 13

EE 3033 FEC CO4 M4 03 STAR Q No 2

EE 3033 FEC CO4 M4 03 STAR Q No 11

EE 3033 FEC CO4 M4 03 STAR Q No 10 star-connected inductors take 8 kW at power factor 0.8 460-V,

EE 3033 FEC CO4 M4 03 STAR Q No 9

EE 3033 FEC CO4 M4 03 STAR Q No 5

EE 3033 FEC CO4 M4 03 STAR Q No 1 A star- 8 + j6per phase is connected to a balanced 3-phase 400-V

EE 3033 FEC CO4 M4 03 STAR Q No 4

EE 3033 FEC CO4 M4 03 STAR Q No 8

EE 3033 FEC CO4 M4 03 STAR Q No 6

EE 3033 FEC CO4 M4 03 STAR Q No 3 Three equal star-connected inductors take 8 kW at a power fact0.8

EE 3033 FEC CO4 M4 03 DELTA Q No 12 A 400-V, 50 Hz, delta Connected 25 of 0.15 H, 120 F in series.

EE 3033 FEC M3 03 Q15 Three impedances are connected in parallel, Write the sinusoidal variations

EE 3033 FEC M3 03 Q14 20 50 Hz supply.4.3 A and the resistive current is 2.7 A. The total 5.8 A

EE 3033 FEC M3 03 Q13 Z1= 15+j12 ohms and Z2= 8-j5 250+j0V, calculate I

EE 3033 FEC M3 03 Q12 A V= 200 ∠53º8' impedances are (12+j16) and (10-j20).

EE 3033 FEC M3 03 Q11 A 5 Ω a0.02 H is arranged in parallel 1 Ω 0.08 H. Find I 100V 50Hz .

EE 3033 FEC M3 03 Q10 Find the reading of the ammeter when the voltmeter across the 3ohm 45 V.

EE 3033 FEC M3 03 Q9 If the voltmeter in Fig reads 60 V, find the reading of the ammeter.

EE 3033 FEC M3 04 Resonance QNo 1 A L = 5.52 mH and R = 10 Ω, to a 100-V, 50-Hz supply. Find C

EE 3033 FEC M3 04 Resonance Q No 2 A 500 kHz and for bandwidth of operation equal to 20 kHz. 5 Ω.

EE 3033 FEC M3 04 Resonance Q No 3 An inductive 2 ohm 0.01 H is connected to a 250-V, 50-Hz.

EE 3033 FEC M3 04 Resonance QNo 4 Find active components 0.1 henry, 8 Ω, 120 µF 240V, 50-Hz s.

EE 3033 FEC M3 04 Resonance QNo 5, R=20 Ω L= 200 µH This combination is in series 8000 Ω.

EE 3033 FEC M3 03 Q7 Z1= (6 - j 8) Z2=(16 + j12) parallel. I= (20+j10), find power

EE 3033 FEC M3 03 Q8 A 15mH series with a parallel 80 Ω resistor and 20 µF capacitor.

EE 3033 FEC M3 03 Q6 10 A flows impedance(2-j5)Ω , (6 +j3) Ω and (3 + j4) Ω. Calculate the current

EE 3033 FEC M3 03 Q5 i_a= 7.07 sin (314t- π/4) i_b = 21.2 sin (314t+π/3) inductive or capacitive.

EE 3033 FEC M3 03 Q3 Refer to the circuit of Fig. and determine the resistance and reactance

EE 3033 FEC M3 03 Q4 A mercury vapour lamp unit consists of a 25µf condenser in parallel with

EE 3033 FEC M3 03 Q2 Z1=8 + j6and Z2=3 - j4 are in parallel I= 25 Afind the current taken and power

EE 3033 FEC M3 03 Q1 Z1=10+j15 and Z2=6-j8 are connected in parallel. If the total current15 A,

EE-3033-FEC-M2.04Basic equation -Explain resonance and Q-factor in series circuit and solve problems

EE 3033 FEC CO2 M2 04 resonance Q No.10 A series L.C circuit has L = 100 pH, C = 2500 µF and Q = 70.

EE 3033 FEC CO2 M2 04 resonance Q No.5 An R-L-C series 1000Ω,100mH 10µµF. If 100 V find resonance f

EE 3033 FEC CO2 M2 04 resonance Q No.4 A 20Ω 25-V, 400-Hz, the current maximum 0.5 A and 150 V.

EE 3033 FEC CO2 M2 04 resonance Q No.9 A 25 Ω Q=10 at a resonant frequency of 10 kHz 100

EE 3033 FEC CO2 M2 04 resonance Q No.7 An R-L-C series resonant = 5000/2"π"Hz; 56Ω Q-factor = 25.

EE 3033 FEC CO2 M2 04 resonance Q No.8. 20-V, R = 20 Ω,L = 20 mH andC = 0.5 µF, resonant frequency

EE 3033 FEC CO2 M2 04 resonance Q No. 6 A R = 1000 Ω, L= 100 mH and C = 10. 100 V resonant f.

EE 3033 FEC CO2 M2 04 resonance Q No 3 A L= 50 mH and R=10 Ω, C= 25µF, 200 V RLC SERIES. find f0

EE 3033 FEC CO2 M2 04 resonance Q No 2 For a series R.L.C inductor is variable. v= 200√2 sin100πt.

EE 3033 FEC CO2 M2 04 resonance Q No 1 A coil in series with a pure capacitor. 10 V 10,000 Hz.

Summative Assessment-2

EE 3033 FEC CO2 M2 03 RLC Q No 13 In a circuit, v lag the I by 30°. (a) PF lagging or leading ?

EE 3033 FEC CO2 M2 03 RLC Q No 12 Draw a vector for the circuit R and X the V1, and V2 and I.

EE 3033 FEC CO2 M2 03 RLC Q No 11 series 20 µF 230-V, 50-Hz supply. I= 8A P = 200 W.

EE 3033 FEC CO2 M2 03 RLC Q No 10In Fig. calculate (i) current (ii) voltage drops V1, V2 and V3

EE 3033 FEC CO2 M2 03 RLC Q No 9 A 4.7 H, 20 Ω, a 4-µF 100Ω non- in series to a 100-V, 50-Hzsupply.

EE 3033 FEC CO2 M2 03 RLC Q No 8 A resistance R, an L = 0.01 H and a capacitance C in series.

EE 3033 FEC CO2 M2 03 RLC Q No 7 Z1 and Z2 230-V, 50-Hz 100 W and 60 W 0.5 lag 0.6 leading resp

EE 3033 FEC CO2 M2 03 RLC Q No 6 A e(t) = 100sin314t,R=10, L= 0.0318, C=63.6 µF. expression for i(t)

EE 3033 FEC CO2 M2 03 RLCQ No 5 A R= 20Ω, L= 0.2 H and C= 100 µ F in series 220-V, 50-Hz mains.

EE 3033 FEC CO2 M2 03 RCQ No 10 A C=10 µF 10°. 100 Ω 220-V, 50-Hz line, Find resistance due C

EE 3033 FEC CO2 M2 03 RCQ No 9. A 700W p.f.= 0.707 leading. v= 141.1 sin (314 t + 30°), find element

EE 3033 FEC CO2 M2 03 RCQ No 8. It is 100-W, 120-V lamp at its current rating from a 240-V, 50-Hz.