Monday, June 01, 2009

Kerala Engineering Entrance 2009 Questions (MCQ) on Alternating Current Circuits

Imagination is more important than knowledge
– Albert Einstein

Alternating current circuits were discussed on this site in the post dated 26th October 2006. You will find the post here.

Today we will discuss the questions from alternating currents which appeared in Kerala Engineering Entrance 2009 question paper. Here are the questions:

(1) In an LCR series ac circuit the voltage across L, C and R is 10 V each. If the inductor is short circuited, the voltage across the capacitor would become

(a) 10 V

(b) 20/√2 V

(c) 20√2 V

(d) 10/√2 V

(e) 20 V

Since the voltages across L and C are of the same value, the circuit is at resonance and Lω = 1/ where ω is the angular frequency. Under resonance the voltage across the resistor is equal to the supply voltage. So the supply voltage is 10 V.

When the inductor is short circuited, the LCR circuit reduces to an RC circuit and the supply voltage is the vector sum of the voltages across R and C so that we have

10 = √(VC2 + VR2)

The voltage drops VC and VR across the capacitor and resistor respectively are equal in value since Lω = 1/Cω = R as judged from the equal voltage drops across L, C and R in the series LCR circuit. Therefore, from the above equation VC = VR = 10/√2 V

(2) A transformer of efficiency 90% draws an input power of 4 kW. An electrical appliance connected across the secondary draws a current of 6 A. The impedance of the device is

(a) 60 Ω

(b) 50 Ω

(c) 80 Ω

(d) 100 Ω

(e) 120 Ω

Efficiency, η = Po/Pi with usual notations. Therefore, the output power is given by

Po = η Pi = 0.9 ×4000 W = 3600 W.

In the problem nothing is mentioned about the power factor cos φ. At your level the load is usually assumed to be effectively resistive so that the power factor may be assumed to be unity. Therefore we have

Po = Io2R where R is the load resistance (or, the load impedence in this case).

This gives R = Po /Io2 = 3600 /62 = 100 Ω.

(3) The impedance of an R-C circuit is Z1 for a frequency f and Z2 for frequency 2f. Then Z1/Z2 is

(a) between 1 and 2

(b) 2

(c) between ½ and 1

(d) ½

(e) 4

The impedance of an R-C circuit is √(R2 + 1/C2ω2).

Since the angular frequency ω =f we have

Z1/Z2 = [√(R2 + 1/C2ω2)] /[√(R2 + 1/4C2ω2)]

For large value of ω the value of Z1/Z2 tends to 1.

For small value of ω the value of Z1/Z2 tends to 2.

Therefore Z1/Z2 lies between 1 and 2.

(4) When a circular coil of radius 1 m and 100 turns is rotated in a horizontal uniform magnetic field, the peak value of emf induced is 100 V. The coil is unwound and then rewound into a circular coil of radius 2 m. If it is rotated now, with the same speed, under similar conditions, the new peak value of emf developed is

(a) 50 V

(b) 25 V

(c) 100 V

(d) 150 V

(e) 200 V

When a coil having N turns and area A rotates in a uniform magnetic field B with angular velocity ω, the peak value of the emf (Vmax) induced in the coil is given by

(Vmax) = NABω

The area of the coil in the second case is 4A since the radius is doubled. But the number of turns will be N/2 in the second case since the length of the wire required for a turn is doubled.

If V1 and V2 are the peak values of the emf in the two cases, we have

V1/V2 = 100 /V2 = NABω / [(N/2)×(4A)]

Or, 100 /V2 = ½ from which V2 = 200 volt.

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