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Question : 36
Total: 37
(a) Show that an ideal conductor does not dissipate power in an a.c. circuit.
(b) The variation of inductive reactance( X L ) of an inductor with the frequency ( f ) of the a.c. source of 100 V and variable frequency is shown in the figure.
(i) Calculate the self-inductance of the inductor.
(ii) When this inductor is used in series with a capacitor of unknown value and a resistor of10 Q at 300 s − 1 , maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.
OR
(a) A conductor of length 'l ' is rotated about one of its ends at a constant angular speed ' ω ' in a plane perpendicular to a uniform magnetic field B. Plot graphs to show variations of the emf induced across the IP variations of the emf induced across the ends of the conductor with (i) angular speed ω and (ii) length of the conductor l .
(b) Two concentric circular loops of radii1 cm and 20 cm are placed coaxially.
(i) Find mutual inductance of the arrangement.
(ii) If the current passed through the outer loop is changed at a rate of5 A ∕ ms , find the emf induced in the inner loop. Assume the magnetic field on the inner loop to be uniform.
(b) The variation of inductive reactance
(i) Calculate the self-inductance of the inductor.
(ii) When this inductor is used in series with a capacitor of unknown value and a resistor of
OR
(a) A conductor of length '
(b) Two concentric circular loops of radii
(i) Find mutual inductance of the arrangement.
(ii) If the current passed through the outer loop is changed at a rate of
Solution:
(a) Power dissipation = P = V RMS I RMS cos ϕ
cos ϕ =
.
For ideal inductorR = 0
∴ cos ϕ = 0
∴ P = V RMS I RMS cos ϕ = 0
Thus, ideal inductor does not dissipate power in an ac circuit.
(b) (i) Inductive reactance= X L = 2 π fL
∴ L =
From graph f = 100 Hz
X L = 20 Ω
∴ L =
=
= 0.032 H = 32 mH
(ii) Power dissipation is maximum when
2 π fL =
f = 300 s − 1
L = 0.032 H
2 π f L =
Or2 π × 300 × 0.032 =
∴ C = 8.8 × 10 − 6 F = 8.8 µ F
OR
(a) (i)e = 1 ∕ 2 B ω L 2
So,e ∝ ω
So, the graph is :
(ii)e = 1 ∕ 2 B ω L 2
So,e ∝ L 2
So, the graph is :
(b) (i)M =
µ 0 = 4 π × 10 − 7
r = 1 cm = 0.01 m
R = 20 cm = 0.2 m
Putting the values
M =
= 0.00986 × 10 − 7 H = 9.98 × 10 − 4 H
(ii)| e | = M
Or, e | = 9.98 × 10 − 4 H × 5000 A ∕ s
∴ Induced emf = | e | = 9.98 × 0.5 = 4.99 V
For ideal inductor
Thus, ideal inductor does not dissipate power in an ac circuit.
(b) (i) Inductive reactance
(ii) Power dissipation is maximum when
Or
OR
(a) (i)
So,
So, the graph is :
(ii)
So,
So, the graph is :
(b) (i)
Putting the values
(ii)
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