Wednesday, May 09, 2007

Kerala Govt Engineering Entrance 2007-Questions on Alternating Currents

Kerala Government Engineering Entrance 2007 question paper contained four questions from alternating currents out of 72 questions in the physics part. Here are the questions:
(1) A square coil of side 25 cm having 1000 turns is rotated with uniform speed in a magnetic field about an axis perpendicular to the direction of the field. At an instant t the e.m.f. induced in the coil is e = 200 sin100πt. The magnetic induction is
(a) 0.50 T (b) ) 0.02 T (c) ) 10–3 T (d) ) 0.1 T (e) ) 0.01 T
If you remember the production of alternating voltage when a plane coil rotates in a magnetic field, you will be able to answer this question in no time. The induced alternating e.m.f. is
e = naBω sin (ωt) and the maximum emf induced is naBω where ‘n’ is the number of turns, ‘a’ is the area (of the coil), B is the magnetic induction and ‘ω’ is the angular speed. So, we have
1000×(25×10–2)2×B×100π = 200. [The value of ‘ω’ is obtained from the form of the equation for the emf].
This gives B = 0.01 T
(2) When a d.c. voltage of 200 V is applied to a a coil of self inductance 2√3/π H, a current of 1 A flows through it. But by replacing d.c. source with a.c. source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of a.c. supply is
(a) 100 Hz (b) 75 Hz (c) 60 Hz (d) 30 Hz (e) 50 Hz
In an LR circuit the direct current is limited by resistance only. On applying a direct voltage of 200V, the current flowing is 1 A which means that the resistance of the coil is 200 Ω.
When the a.c. source is connected, the current is given by I = V/Z where the impedance Z is given by Z = √(R2+L2 ω2)
Therefore 0.5 = 200/√[2002+(2√3/π)2 ω2] = 200/√[2002+(2√3/π)2 (2πf)2], from which f = 50 Hz.
(3) In a LR circuit, the value of L is (0.4/π) henry and the value of R is 30 Ω. If in the circuit, an alternating emf of 200 V at 50 cycles per second is connected, the impedance of the circuit and the current will be
(a) 11.4 Ω, 17.5 A (b) 30.7 Ω, 6.5 A (c) 40.4 Ω, 5 A
(d) 50 Ω, 4 A (e) 35 Ω, 6.5 A
This question is quite simple and straight forward.
Impedance, Z = √(R2+L2 ω2) = √[(302+(0.4/π)2 (100π)2)] since ω = 2πf =2π×50 = 100π.
This gives Z = 50 Ω.
Current I = V/Z = 200/50 = 4 A.
(4) A transformer has an efficiency of 80%. It is connected to a power input of 5 kW at 200 V. If the secondary voltage is 250 V, the primary and secondary currents are respectively
(a) 25 A, 20 A (b) 20 A, 16 A (c) 25 A, 16 A (d) 40 A, 25 A (e) 40 A, 16 A
This question also is quite simple and can be worked out within a minute.
Primary current IP = (5000 watts)/(200 volts) = 25 A.
Secondary current = (5000×0.8 watts)/(250 volts) = 16 A.
[Note that the output power from the secondary side is efficiency times the input power].

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