(a) Show that an ideal inductor does not dissipate power in an ac circuit. (b) The variation of inductive reactance (XL) of an inductor with the frequency (f) of the ac source of 100 V and variable frequency is shown in fig.
(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 10 Ω 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 variation 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 radius 1 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 of 5 A/ms, find the emf induced in the inner loop. Assume the magnetic field on the inner loop to be uniform.
[2020 | Delhi | Section - D | SET - 1 SET - 2 SET - 3, Q.36 Q.37 Q.35
Solution :
i) We know
ε = ε₀ sinωt
I = I₀ sin (ωt-π/2)
Instantaneous Power Supplied
P = εI
P = [ε₀ sinωt]×[I₀ sin (ωt-π/2)]
P = ε₀I₀sinωt sin (ωt-π/2)
P = -[ε₀I₀/2] sin2ωt
Now, average Power over complete cycle will be
<P> = <-[ε₀I₀/2] sin2ωt>
<P> = -[ε₀I₀/2]<sin2ωt>
<P> = -[ε₀I₀/2]×0
[Average value of sine function over a complete cycle is zero]
<P> = 0
Thus the average power dissipated in a pure inductive circuit is zero. [2 Marks]
ii) V = 100 V
XL = 2πfL
XL/2πf = L
L = 40/(2π×200)
L = 7/220
L = 0.032 H
ii) Power dissipation will be maximum at resonance.
Here
R = 10Ω
f = 300 s-1
f =1/2π√(LC)
C =1/(4π2Lf2)
C =1/[4×(22/7)2×0.032×(300)2]
C =49/[5575680]
C = 8.788 × 10-6 F
C = 8.788µF
C = 8.8µF
OR
a) ε = ½Bl2ω
Plot the graph by using the above equation.
b) (i) Mutal Inductance
r1 = 20 cm = 0.2 m
r2 = 1 cm = 0.01 m
Consider the current I is flowing through larger coil, then magnetic flux linked with the smaller coil will be
Φ = MI
BA = MI
[µ₀I/2r1]A = MI
M = µ₀A/2r1
M = µ₀(2πr22)/2r1
M = [(4π×10-7)×(2π(0.01)2)]/2(0.2)
M = π2×10-10 H
M = 9.8596×10-10 H
M = 10×10-10 H
M = 10-9 H
ii)
dI/dt = 5A/ms
dI/dt = 5000 A/s
ε = -MdI/dt
ε = -(10-9)×5000
ε = -5×10-6 V
