(a) Show that an ideal conductor does not ...

CBSE Class 12 Physics 2020 Delhi Set 1 Paper

Question : 36 of 37

Marks: +1, -0

(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\,\mathrm{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 of

10 Q

300\,\mathrm{s}^{-1}

, maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.

(a) A conductor of length '

l

' is rotated about one of its ends at a constant angular speed '

\omega

' 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

\omega

and (ii) length of the conductor

l

(b) Two concentric circular loops of radii

1\,\mathrm{cm}

and

20\,\mathrm{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\,\mathrm{A}/\mathrm{ms}

, find the emf induced in the inner loop. Assume the magnetic field on the inner loop to be uniform.

Solution:

(a) Power dissipation

= P = V_{\mathrm{RMS}} I_{\mathrm{RMS}} \cos \phi

\cos \phi = \frac{R}{Z}.

For ideal inductor

R = 0

\;\therefore\;\; \cos \phi = 0

\;\therefore\;\; P = V_{\mathrm{RMS}} I_{\mathrm{RMS}} \cos \phi = 0

Thus, ideal inductor does not dissipate power in an ac circuit.

(b) (i) Inductive reactance

= X_L = 2\pi fL

\therefore\;\; L = \frac{X_L}{2\pi f}

\;\;\text{ From graph }\;\; f = 100\,\mathrm{Hz}

X_L = 20\,\Omega

\therefore L = \frac{X_L}{2\pi f} = \frac{20}{2\pi \times 100} = 0.032\,\mathrm{H} = 32\,\mathrm{mH}

(ii) Power dissipation is maximum when

2\pi fL = \frac{1}{2\pi f C}

f = 300\,\mathrm{s}^{-1}

L = 0.032\,\mathrm{H}

2\pi f L = \frac{1}{2\pi f C}

2\pi \times 300 \times 0.032

= \frac{1}{2\pi \times 300 \times C}

\therefore C = 8.8 \times 10^{-6}\,\mathrm{F} = 8.8\,\mu\mathrm{F}

(a) (i)

e = \frac{1}{2} B \omega L^2

So,

e \propto \omega

So, the graph is :

(ii)

e = \frac{1}{2} B \omega L^2

So,

\;\; e \propto L^2

So, the graph is :

(b) (i)

M = \frac{\mu_0 \pi r^2}{2R}

\mu_0 = 4\pi \times 10^{-7}

r = 1\,\mathrm{cm} = 0.01\,\mathrm{m}

R = 20\,\mathrm{cm} = 0.2\,\mathrm{m}

Putting the values

M = \frac{4\pi \times 10^{-7} \times \pi (0.01)^2}{2 \times 0.2}

= 0.00986 \times 10^{-7}\,\mathrm{H} = 9.98 \times 10^{-4}\,\mathrm{H}

(ii)

|e| = M \frac{\Delta i}{\Delta t}

\;\;\text{ Or, }\;\; e| = 9.98 \times 10^{-4}\,\mathrm{H} \times 5000\,\mathrm{A}/\mathrm{s}

\;\therefore\;\text{ Induced emf }\; = |e| = 9.98 \times 0.5

= 4.99\,\mathrm{V}

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