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New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

In steady state, capacitor branch is open.
V_AB = 2Ω (1A) + 2Ω (3A) = 8V.

 

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

V_PQ * (49/100). 1.02 = V_PQ * 0.49. V_PQ ≈ 2.08V.
Potential gradient = 2.08V/100cm = 0.0208 V/cm.

New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

m = I (abk + abj)
|m| = Iab√2
Direction ⇒ (j+k)/√2

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

ρ? = 98 * 10?
ρ? = 2.65 * 10?
ρc = 1.724 * 10?
ρT = 5.65 * 10?

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Figure of Merit = C = i/θ
= C = (6 * 10? ³)/2 = 3 * 10? ³ Am²

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

r in   = l 1 - l 2 l 2 R ext   = 60 500 * 10

r = 6 5 = 1.2 Ω

n = 12

 

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

 

Potential gradient = 5 1000 = V P 1200

V P = 6 V

and  R P = V P l = 6 60 * 10 - 3 = 100 Ω

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

  I = 1 2.5 = 0.4 A

I = I 2 = 0.2 A

 

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