Alternating Current

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Question 1

A circuit has a resistance of 12 Ω and an impedance of 15 Ω. The power factor of the circuit will be

  • 0.8

  • 0.4

  • 1.25

  • 0.125

Solution

A.

0.8

cosϕ space equals space straight R over straight Z space equals space 12 over 15 space equals 4 over 5 space equals space 0.8
Question 2

A conductor lies along the z-axis at -1.5 ≤ z< 1.5 m and carries a fixed current of 10.An in-az direction (see figure).
B = 3.0 x 10-4 e-0.2x ay T, find the power required to move the conductor at constant speed to x = 2.0 , y = 0  in 5 x10-3 s. Assume parallel motion along the x axis.

  • 1.57 W

  • 2.97 W

  • 04.85 W

  • 29.7 W

Solution

B.

2.97 W

When force exerted on a current carrying conductor
Fext = BIL
average space power space equals space fraction numerator Work space done over denominator TIme space taken end fraction
straight P space equals 1 over straight t integral subscript 0 superscript 2 straight F subscript ext. end subscript space dx space equals space 1 over straight t integral subscript 0 superscript 2 straight B space left parenthesis straight x right parenthesis IL space dx
space equals space fraction numerator 1 over denominator 5 space straight x space 10 to the power of negative 3 end exponent end fraction integral subscript 0 superscript 2 space 3 space straight x space 10 to the power of negative 4 end exponent straight e to the power of negative 0.2 end exponent space straight x space 10 space straight x space space 3 dx
space equals space 9 open square brackets 1 minus blank over straight e to the power of 0.4 end exponent close square brackets
space equals space 2.967 space almost equal to 2.97 space straight W

Question 3

A current I flows along the length of an infinitely long, straight, thin walled pipe. Then 

  • the magnetic field is zero only on the axis of the pipe

  • the magnetic field is different at different points inside the pipe

  • the magnetic field at any point inside the pipe is zero

  • the magnetic field at all points inside the pipe is the same, but not zero

Solution

C.

the magnetic field at any point inside the pipe is zero

Question 4

A current I flows in an infinitely long wire with cross-section in the form of a semicircular ring of radius R. The magnitude of the magnetic induction along its axis is

  • fraction numerator straight mu subscript 0 straight I over denominator 2 πR end fraction

  • fraction numerator straight mu subscript 0 straight I over denominator 2 πR end fraction
  • fraction numerator straight mu subscript 0 straight I over denominator 4 πR end fraction
  • fraction numerator straight mu subscript 0 straight I over denominator straight pi squared straight R end fraction

Solution

D.

fraction numerator straight mu subscript 0 straight I over denominator straight pi squared straight R end fraction straight v equals space straight I over πR
dB space equals space open parentheses fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction close parentheses fraction numerator 2 straight I over denominator straight R end fraction
straight I space equals space λRdθ
therefore space straight B space equals space integral subscript negative straight pi divided by 2 end subscript superscript straight pi divided by 2 end superscript space dB space cos space straight theta
space equals space fraction numerator straight mu subscript 0 straight lambda over denominator 2 straight pi end fraction integral subscript negative straight pi divided by 2 end subscript superscript straight pi divided by 2 end superscript space cos space θdθ
space equals space fraction numerator straight mu subscript 0 straight lambda over denominator straight pi end fraction space equals space fraction numerator straight mu subscript 0 straight I over denominator straight pi squared straight R end fraction
Question 5

A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is:

  • 1.73 V/m

  • 2.45 V/m

  • 5.48 V/m

  • 7.75 V/m

Solution

B.

2.45 V/m

Consider the LED as a point source of light.
Let power of the LED is P

Intensity at r from the source
straight I space equals space fraction numerator straight P over denominator 4 πr squared end fraction ... (i)
As we know that,
straight I space equals space 1 half straight epsilon subscript straight o straight E subscript 0 superscript 2 straight c .... (ii)
From eqs. (i) and (ii) we can write
fraction numerator straight P over denominator 4 πr squared end fraction space equals space 1 half space straight epsilon subscript straight o straight E subscript 0 superscript 2 straight c
straight E subscript 0 superscript 2 space equals space fraction numerator 2 straight P over denominator 4 πε subscript 0 straight r squared straight c end fraction space equals space fraction numerator 2 space straight x space 0.1 space straight x space 9 space straight x space 10 to the power of 9 over denominator 1 space straight x space 3 space straight x space 10 to the power of 8 end fraction
straight E subscript 0 superscript 2 space equals space 6
straight E subscript 0 space equals space square root of 6 space equals space 245 space straight V divided by straight m

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