Alternating Current
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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
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3.6 A
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2.8 A
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2.5 A
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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:
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qvR/2
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qVR2
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qVR2/2
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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
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0.56
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0.64
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0.80
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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:
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500
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1000
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1250
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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
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putting in parallel a resistance of 240 Ω
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putting in series a resistance of 15 Ω
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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
Mock Test Series