Physics Current Electricity

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New answer posted

3 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

3 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  R i = ρ l A
R f = ρ ( 1 . 2 5 l ) ( A / 1 . 2 5 ) = ( 1 . 2 5 ) 2 * ρ l A

R f = 1 . 5 6 2 5 * R i

R f R l R l * 1 0 0 = 5 6 . 2 5 %

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

d q d t = t = 2 0 t + 8 t 2

0 q d q = 0 1 5 ( 2 0 t + 8 t 2 ) d t

q = 2 0 * 1 5 2 2 + 8 . 1 5 3 3 = 1 1 2 5 0 C

New answer posted

3 weeks ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

D 1 D 3 } Forward biased offer zero Resistance

D2} Reversed biased offers Infinite Resistance

I = v R = 1 0 1 0 = 1 A m p

 

New answer posted

3 weeks ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Volume = constant

A l = C  

A d l + l d A = 0  

d l l + d A A = 0  

d R R * 1 0 0 = ( d l l d A A ) * 1 0 0  

= 0.4 - (-0.4)

= 0.8 %

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

According to Kirchhoff's Law, we can write

-20 + 2000I + 600 * 5I = 0 I = 2 0 5 0 0 0 A  

Reading of voltmeter = 2000I = 2000 * 2 0 5 0 0 0 = 8 v o l t  

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

According to question, we can write

2 V 2 + 2 r = V 2 + r 2

2 + 2 r = 4 + r r = 2 Ω

New answer posted

3 weeks ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Using whetstone

R 3 = 4 6 R = 2 Ω

R e q u = 5 4 1 5 Ω

i = 3 6 5 4 1 5 = 1 0 A

New answer posted

3 weeks ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

At null point:

1 5 4 3 + 2 = R 1 0 0 4 3

R = 1 9 Ω

New answer posted

3 weeks ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

According to Kirchhoff's Law, we can write

 

-20 + 2000I + 600 * 5I = 0  I = 2 0 5 0 0 0 A  

Reading of voltmeter = 2000I = 2000 *  2 0 5 0 0 0 = 8 v o l t  

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