Class 12th

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New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

i = V r 1 - e - R t / L = 4 1 - e - 500 t

At t = ∞

i 1 = 4 A

at t = 40 s

i 2 = 4 1 - e - 20000

= 4 1 - 1 e 2 10000

= 4 1 - 1 ( 7.389 ) 10000

i 1 i 2 is slightly greater than 1 .

New answer posted

a year ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 14 Views

V
Vishal Baghel

Contributor-Level 10

I010=I0cos? 2? \frac {I_0} {10} = I_0 \cos^2 \theta cos? ? =110=0.31<12which is 0.707\cos \theta = \frac {1} {\sqrt {10} = 0.31 < \frac {1} {\sqrt {2} \quad \text {which is 0.707}

So,

? >45? and90? ? ? <45? \theta > 45^\circ \quad \text {and} \quad 90^\circ - \theta < 45^\circ

so only one option is correct i.e. 18.4°

Angle rotated should be

=90? ? 71.6? =18.4? = 90^\circ - 71.6^\circ = 18.4^\circ

Answer: (A) 18.4°

New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

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R
Raj Pandey

Contributor-Level 9

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New answer posted

a year ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution
 

 

New answer posted

a year ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

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