Wired Faculty
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A 220 V input is supplied to a transformer. The output circuit draws a current 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
3.6 A
2.8 A
2.5 A
5.0 A
D.
5.0 A
Efficiency is defined as the ratio of output power and input power
i.e. 
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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
qvR
A.
qvR/2
As revolving charge is equivalent to a current so
I = qf = q x ω/ 2π
But ω = v/R
Where R is radius of circle and v is uniform speed of charged particle, therefore,
I= qv/2πR
Now, magnetic moment associated with charged particle is given by
μ = IA = I x πR2
μ = qv/2 πR2
= qvR/2
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
0.33
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, 
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
100
C.
1250
AC power gain is ratio of change in output power to the change in input power.
AC power gain
A galvanometer having a coil resistance of 60 Ω shows full-scale deflection when a current of 1.0 A passes through it. It can be converted into an ammeter to read currents up to 5.0 A by
putting in parallel a resistance of 240 Ω
putting in series a resistance of 15 Ω
putting in series a resistance of 240 Ω
putting in parallel a resistance of 15 Ω
D.
putting in parallel a resistance of 15 Ω
To convert a galvanometer to ammeter a small resistance is connected in parallel to the coil of the galvanometer.
Here G1 = 60 Ω, Ig = 1.0 A, I = 5 A
Ig G1 = (I - Ig) S
Putting 15 Ω resistance in parallel.
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Mock Test Series
