– Albert Einstein

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 *Lω = *1/*Cω* where *ω* is the angular frequency. Under resonance the voltage across the resistor is equal to the supply

When the

10 = √(V_{C}^{2 }+ V_{R}^{2})

The _{C} and V_{R} 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 V_{C }= V_{R} = **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 = *P*_{o}*P*_{i} with usual notations. Therefore, the output power is given by

*P*_{o}* P*_{i} = 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

*P*_{o}*I*_{o}^{2}*R* where *R* is the load resistance (or, the load impedence in this case).

This gives *R = P*

_{o}

_{ }/

*I*

_{o}

^{2}

*= 3600 /6*

^{2}= 100 Ω.

**(3**) The impedance of an R-C circuit is *Z*_{1 }for a frequency *f* and *Z*_{2} for frequency 2*f*. Then *Z*_{1}/*Z*_{2} 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 √(R^{2} + 1/C^{2}*ω*^{2}).

Since the angular frequency *ω =* 2π*f* we have

*Z*_{1}/*Z*_{2} = [√(R^{2} + 1/C^{2}*ω*^{2})] /[√(R^{2} + 1/4C^{2}*ω*^{2})]

For *large* value of *ω* the value of *Z*_{1}/*Z*_{2} tends to 1.

For small value of *ω* the value of *Z*_{1}/*Z*_{2} tends to 2.

Therefore *Z*_{1}/*Z*_{2} 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 (*V*_{max})* *induced in the coil is given by

(*V*_{max}) = *NABω*

The area of the coil in the second case is 4*A* 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 *V*_{1} and *V*_{2} are the peak values of the emf in the two cases, we have

*V*_{1}/*V*_{2} = 100 /*V*_{2} = *NABω / *[(*N/*2)×(4*A*)*Bω*]

Or, 100 /*V*_{2} = ½ from which* V*_{2} = 200 volt.

Heavy computation!

ReplyDeleteQuestion 4 under Kerala Engineering Entrance 2009 Questions (MCQ) on Alternating Current Circuits

ReplyDeleteWhere can i find the derivation for induced emf=NABw

If the area vector (which is perpendicular to the plane of the coil, by convention) makes an angle θ with the magnetic field vector, the magnetic flux (φ) linked with the coil is AB cos θ for a single turn coil and is NABcos θ for a coil of N turns. Here θ is the angle turned by the coil in time t so that θ = ωt where ω is the angular velocity of rotation of the coil. Therefore φ = NABcos ωt

ReplyDeleteSince the induced emf is –dφ/dt, the instantaneous value of induced emf is NABω sin ωt

The maximum value of the induced emf is therefore NABω

Thankyou,Sir.

ReplyDelete