Class 12th

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New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

i ( z e n e r − m a x ) = 2 5 m A                                                                

 20 – imax R – 8 = 0

imax R = 12At minimum zener current ( μ A ) : −  

2 0 − i m i n R − i m i n R L = 0  

  R R L = 1 2 8 = 3 2

l m i n R = 1 2

  i m i n R L = 8

At maxm zener current –

2 0 − i m a x R − 8 = 0  

i L = O { a s     i z     m a x m = 2 5 m A }             

imaxR = 12v

25 * 10-3 R = 12

R = 1 2 * 1 0 3 2 5 = 1 2 * 4 0 = 4 8 0 Ω

 

New answer posted

a year ago

0 Follower 30 Views

V
Vishal Baghel

Contributor-Level 10

C e f f = [ ε 0 ( 7 * 4 ) 4 / 1 0 + 5 ε 0 ( 1 * 4 ) 4 / 1 0 ] * 1 0 − 2

C e f f = 1 . 2 ε 0

Energy = 1 2 C e f f V 2

1 2 ( 1 . 2 ε 0 ) ( 2 0 ) ( 2 0 ) = 2 4 0 ε 0

New answer posted

a year ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Modulating signal 2sin (6.28 * 106)t

Carrier signal  4 sin (12.56 * 109)t

New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

q = C V 1 0 0 Ω

= ( 1 . 1 * 1 0 − 6 ) ( 1 0 R + r R )

= 1 . 1 * 1 0 − 6 ( 1 0 1 1 0 * 1 0 0 )

= 1 0 μ C

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

? = B → . A →

= ( 3 t 3 j ^ + 3 t 2 k ^ ) . ( π ( 1 ) 2 k ^ )

? = 3 t 2 π

E i n d = | d ? d t | = 6 t π

at t = 2, Eind = 12

New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

δ = A ( μ y − 1 ) − A 1 ( μ y 1 − 1 )

= 6 ( 1 . 5 − 1 ) − 5 ( 1 . 5 5 − 1 )

= 1 4

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Independent of are a is case of uniform wire

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

P = 10% of 110W = 11W

(Visible)

I 1 − I 2 = ρ 4 π ( 1 r 1 2 − 1 r 2 2 )

= 1 1 4 * 2 2 7 ( 1 1 − 1 2 5 )

7 8 * ( 2 4 2 5 ) = a * 1 0 − 2 W / m 2

2 8 * 2 4 8 * 1 0 − 2 = a * 1 0 − 2

=84 * 10-2 = a * 10-2 = a = 84

 

New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

take z = x + iy

z2+z¯=0

⇒x2−y2+x+i  2xy−yi=0

⇒x2−y2+x=0  and  y (2x−1)=0

if y = 0 x = 0, 1

i f     x = 1 2 ⇒ y = ± 3 2

Σ ( R e ( z ) + l m ( z ) ) = ( 0 − 1 + 1 2 + 1 2 ) + ( 0 + 0 + 3 2 − 3 2 ) = 0

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Let A =  [abcdef9hi]

Now ATA

trace will be a2+b2+c2+d2+e2+f2+92+h2=6

total ways = 9C6.26.1.1.1 = 5376

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