The primary and secondary coil of a transformer have 50 and 1500 turns respectively. If the magnetic flux Φ linked with the primary coil is given Φ = Φo +4t, where Φ is in weber, t is time is second and Φo is a constant the output voltage across the secondary coil is:
90 V
120 V
220 V
220 V
A charged particle (charge q ) is moving n a circle of radius R with uniform speed v. The associated magnetic moment μ is given by:
qvR/2
qVR2
qVR2/2
qVR2/2
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is:
500
1000
1250
1250
The core of a transformer is laminated because
energy losses due to eddy currents may be minimised
the weight of the transformer may be reduced
rusting of the core may be prevented
rusting of the core may be prevented
A coil of inductive reactance 31 Ω has a resistance of 8Ω. It is placed in series with a condenser of capacitative reactance 25Ω. The combination is connected to an a.c. soruce of 110 V. The power factor of the circuit is
0.56
0.64
0.80
0.80
C.
0.80
Power factor is the ratio of resistance and impedance of a.c. circuit.
Power factor of a.c. circuit is given by
...(i)
where R is resistance employed and z the impedance of the circuit.
...(ii)
Eqs. (i) and (ii) meet to give,
...(iii)
Given,
In an electromagnetic wave in free space the root mean square value of the electric field is E rms= 6 V/m. The peak value of the magnetic field is
1.41 × 10–8 T
2.83 × 10–8 T
0.70 × 10–8 T
0.70 × 10–8 T
An inductor 20 mH, a capacitor 100 µF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is
0.79 W
0.43
1.13
2.74 W
An AC supply gives 30Vrms which is fed on a pure resistance of 10 ohm. the power dissipated in this is
90watt
45watt
90.2watt
180watt
A transmitting station transmits radiowaves of wavelength 360 m. Calculate the inductance of coil required with a condenser of capacity 1.20 μF in the resonant circuit to receive them
3.07×10-8 H
2.07×10-8 H
4.07×10-8 H
6.07×10-8 H
The peak value of alternating current is ampere. The mean square value of current will be
5 A
2.5 A
A
None of the above