A loss free transformer has 500 turns on its primary winding and 2500 in secondary. The meters of the secondary indicate 200 volts at 8 amperes under these conditions. The voltage and current in the primary is 

1. 100 V, 16 A

2. 40 V, 40 A

3. 160 V, 10 A

4. 80 V, 20 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

A power transformer is used to step up an alternating e.m.f. of 220 V to 11 kV to transmit 4.4 kW of power. If the primary coil has 1000 turns, what is the current rating of the secondary ? Assume 100% efficiency for the transformer 

1. 4 A

2. 0.4 A

3. 0.04 A

4. 0.2 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

The peak value of an alternating e.m.f. E is given by $E=E_0\cos \omega t$ is 10 volts and its frequency is 50 Hz. At time $t=\frac{1}{600}sec$, the instantaneous e.m.f. is

1. 10 V

2. $5\sqrt{3}V$

3. 5 V

4. 1 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