The core of a transformer is laminated to reduce energy losses due to  

1. Eddy currents

2. Hysteresis

3. Resistance in winding

4. None of these

An electric motor operating on a 60 V dc supply draws a current of 10 A. If the efficiency of the motor is 50%, the resistance of its winding is 

1. 3Ω

2. 6Ω

3. 15Ω

4. 30Ω

A step-down transformer is connected to 2400 volts line and 80 amperes of current is found to flow in output load. The ratio of the turns in primary and secondary coil is 20 : 1. If transformer efficiency is 100%, then the current flowing in primary coil will be 

1. 1600 A

2. 20 A

3. 4 A

4. 1.5 A

A transformer connected to 220 Volt line shows an output of 2 A at 11000 Volt. The efficiency is 100%. The current drawn from the line is:

1. 100 A

2. 200 A

3. 22 A

4. 11 A

The primary winding of transformer has 500 turns whereas its secondary has 5000 turns. The primary is connected to an ac supply of 20 V, 50 Hz. The secondary will have an output of 

1. 200 V, 50 Hz

2. 2 V, 50 Hz

3. 200 V, 500 Hz

4. 2 V, 5 Hz

A step-down transformer is connected to main supply 200V to operate a 6V, 30W bulb. The current in primary is 

1. 3 A

2. 1.5 A

3. 0.3 A

4. 0.15 A

A transformer has 100 turns in the primary coil and carries 8 A current. If input power is one kilowatt, the number of turns required in the secondary coil to have 500V output will be 

1. 100

2. 200

3. 400

4. 300

The potential difference V and the current i flowing through an instrument in an ac circuit of frequency f are given by $V=5\cos \omega t$ volts and I = 2 sin ωt amperes (where ω = 2πf). The power dissipated in the instrument is 

1. Zero

2. 10 W

3. 5 W

4. 2.5 W

A generator produces a voltage that is given by V = 240 sin 120 t, where t is in seconds. The frequency and r.m.s. voltage are 

1. 60 Hz and 240 V

2. 19 Hz and 120 V

3. 19 Hz and 170 V

4. 754 Hz and 70 V

If a current I given by $I_0\sin \left(\omega t-\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)$ flows in an ac circuit across which an ac potential of $E=E_0\sin \omega t$ has been applied, then the power consumption P in the circuit will be 

1. $P=\frac{E_0{I}_0}{\sqrt{2}}$

2. $P=\sqrt{2}{E}_0{I}_0$

3. $P=\frac{E_0{I}_0}{2}$

4. P = 0