Physics Current Electricity

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

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

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

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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

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V
Vishal Baghel

Contributor-Level 10

Let us assume the potential at A = V_A = 0.
Now at junction C, according to KCL
i? + i? = i?
1 A + i? = 2 A


i? = 2 A
Now analyse potential along ACDB
V_A + 1 + i? (2) - 2 = V_B
0 + 1 + 2 (1) - 2 = V_B
V_B = 3 - 2
V_B = 1amp.

New answer posted

a month ago

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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

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R
Raj Pandey

Contributor-Level 9

10 R 10 + R * 12 = 15 * 4 on solving

R = 10 Ω

New answer posted

a month ago

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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

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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

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V
Vishal Baghel

Contributor-Level 10

  I = 1 2.5 = 0.4 A

I = I 2 = 0.2 A

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Total power is ( 15 * 45 ) + ( 15 * 100 ) + ( 15 * 10 ) + ( 2 * 1000 ) = 4325 W .

So, current is 4325 220 = 19.66 A

Answer is 20 A m p .

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