Physics Current Electricity

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

ε? = 250 * Potential Gradient
ε? + ε? = 400 * Potential Gradient
(ε? )/ (ε? + ε? ) = 250/400 = 5/8
By solving above
ε? /ε? = 5/3

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

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

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

i = 6 4 2 + 8 = 0 . 2 A

V x + 4 + 0 . 2 * 8 = V Y

V Y V X = 5 . 6 V

 

New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Potential difference across 2k Ω  is 5V, thus current through it,

  i = 5 2 * 1 0 3 = 2 5 * 1 0 4 A .          

New answer posted

a month ago

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V
Vishal Baghel

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

a month ago

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P
Payal Gupta

Contributor-Level 10

W = q Δ V = 2 0 * 1 5 = 3 0 0 J

New answer posted

a month ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

Given circuit can be re-drawn and it becomes case of balanced Wheatstone bride

R A B = ( 2 R ) ( 2 R ) 2 R + 2 R = R

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

radius r = 0.5 mm

σ = 5 * 1 0 7 s / m

E = 1 0 * 1 0 3 V / m = 1 0 2 V / m

As we know

J = σ E

J = i A = σ E

i = σ E A = 5 * 1 0 7 * 1 0 2 * π * ( 0 . 5 * 1 0 3 ) 2 = 1 . 2 5 π * 1 0 1 A = 1 2 5 π m A = x 3 π m A

 x = 5

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

E 1 E 2 = 3 8 0 7 6 0 = 1 2                                                                                                      

New answer posted

a month ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

i2 = 2i1

i1 + i2 = 6

i1 = 2A.

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