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New answer posted

a week ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Final resistance ( R ) = V I = 8 0 Ω

then, R = R0 (1 + µDT)

80 = 60 (1 + 2 * 10–4 DT)

DT = 1666.67

T – 27

T = 1693.66

= 1694°C

New answer posted

a week ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

i = 7 3 . 5 k + 0 . 9 k Ω = 7 4 . 4 k

V 0 = i * 7 0 0 Ω = 7 4 . 4 k * 7 k = 9 . 9 4 . 4 = 1 . 1 V

New answer posted

a week ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

Req = r

P = V 2 r = 4 2 = 2 W

New answer posted

a week ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

p = i2R

P' = 0.64 i2R

New answer posted

a week ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a week ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

i = neAVd

 

New answer posted

a week ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a week ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a week ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

01 * 104 Ω ± 5% = 10 * 103 Ω ± 5%

New answer posted

2 weeks ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

When wire is stretched uniformly

R R

R R ' = l n l 2 = 1 n 2  

R ' = n 2 R

When this elongated wire is cut into 5 parts, resistance of each part  

Effective resistance between A and C = n 2 R 5  

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