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New answer posted

a year ago

0 Follower 16 Views

V
Vishal Baghel

Contributor-Level 10

IgRc = (I - Ig) (S? + S? + S? )
1mA * 10 = 9mA (S? + S? + S? )

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

i series   = E R series  

⇒ i series   = E 10 R

i parallel   = E R Parallel  

= E R / 10

i parallel   = n * i series  

⇒ 10 E R = n E 10 R

⇒ n = 100

New answer posted

a year ago

0 Follower 10 Views

R
Raj Pandey

Contributor-Level 9

R = R 0 ( 1 + α Δ T )

⇒ 6.8 = 2 [ 1 + α * ( 80 - 0 ) ] ⇒ α = 3.4 - 1 80 = 0.03 = 3 * 10 - 2 ? C - 1

New answer posted

a year ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

V = ir
i = V/r
i = (1/r) [v is same for r? & r? ]
i? /i? = r? /r?
i? = (r? / (r? +r? )i?
i? /i? = r? / (r? +r? )

New answer posted

a year ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 37 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

E 1 E 2 = ? l 1 ? l 2

1.5 2.5 = 36 l 2 ⇒ l 2 = 36 * 5 3 = 60 c m

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

R = R 0 ( 1 + α Δ T )

⇒ 6.8 = 2 [ 1 + α * ( 80 - 0 ) ] ⇒ α = 3.4 - 1 80 = 0.03 = 3 * 10 - 2 ? C - 1

New answer posted

a year ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

B ? P = B ? upper wire   ⊗ + B ? semi-circle   ? + B ? lower-wire   ⊗

B P = - μ 0 i 4 π R + μ 0 i 4 R - μ 0 i 4 π R = μ 0 i 4 R 1 - 2 π  pointing away from the page

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Magnetic energy stored in an inductor = 1 2 L I 2

= 1 2 * 4 * 10 - 6 * ( 2 ) 2 = 8 μ J

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