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12th Standard English Medium Physics Subject Electromagnetic Induction and Alternating Current Creative 5 Mark Questions with Solution Part - II

Published on: 18/06/2021

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Std: 12th Standard - Physics

chapter: - Creative Questions

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Questions:

Generic 5

5 x 5 = 25

The instantaneous current from an AC source is given by I = 5 sin 314 t. What is the rms value of the current?

An AC generator consists of a coil of 1000 turns and cross sectional area of 100 cm2, rotating at an angular speed of 100 rpm in a uniform magnetic field of 1.6 x 10^-2 T calculate the maximum emf produced in the coil.

In the circuit diagram shown in figure, R = 10W, L = 5H, E = 20V. I = 2A. This current is decreasing at a rate of -1.0A/S. Find VAB at this instant.

A circular coil of radius 8.0 cm and 20 turns rotated about its vertical diameter with an angular speed of 50 rads in a uniform horizontal magnetic field of magnitude 3 X 10^-2T. Obtain the maximum and average emf induced in the coil if the coil forms a closed loop of resistance 10Ω then calculate the average power loss due to joule heating.

The magnetic flux through a coil perpendicular to the plane is given by Φ = 5t³ +4t² +2t calculate the induced emf through the coil at t = 2S

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Answers Key:

Generic 5

5 x 5 = 25

Given: 1= 5 sin 314 μt ...(1)
We know that I = I_m sin ωt ....(2)
Comparing (1) & (2)
I_m= 5A
= 314
I_rms= \(\frac { { I }_{ m } }{ \sqrt { 2 } } =\frac { 5 }{ \sqrt { 2 } } \) = 3.536A

Given: N = 1000
A = 100 cm² = 10-2 m²
v = 100 rpm = \(\frac{100}{60}\) rps
B = 3.6 x 10^-27
em = ?
em = NBAω = NBA (2πγ)
= 1000 x 3.6 x 10^-2 x 10^-2 x 2 x \(\frac{22}{7}\times\frac{100}{60}\)
e = 5.77 V

Potential difference across inductor is
V_L= L.\(\frac{dI}{dt}\) = 5(-1.0)
V_L = -5.0V
Using kirchoff's II law,
V_A- I_R - V_L - E = V_B
(V_A- V_B) = I_R + V_L + E
i.e V_AB = 20 + (-5.0) + 20
= 35V

Given:
Radius of coil, r = 8.0 cm
= 8 x 10^-2m
N = 20 turns, 0) = 50 rads^-1
To find : B = 3 x 10^-2T, e_m=? e = ?
Resistance R = 10W, Power P = ?
We know, e_m = NABW m
= N(πr²)Bω
em = 20 x \(\frac{22}{7}\) x (8 x 10^-2) x 3 x 10^-2 x 5
e_m= 0.603 V
The average value of emf induced over a full cycle, e_m = 0
I_m = \(\frac{e_m}{R}\) = \(\frac{0.603}{10}\) = 0.603A
Average power dissipated, Pav = \(\frac{e_mI_m}{2}\)
= \(\frac{0.603\times0.0603}{2}\)
Pav = 0.018w

Given:
We know that e = -\(\frac { d\Phi }{ dt } \)
e = \(\frac{d}{dt}\) (5t³ + 4t² + 2t)
e = 15t² + 8t + 2
for t = 2s, e = 15 x (2)²+ 8 (2) + 2
e = 78V