Physics Current Electricity

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

q V = 1 2 m v 2

v = 2 q V m v q m

v H v H e = 1 1 * 4 1 = 2 1

New answer posted

4 months ago

0 Follower 3 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

4 months ago

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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

4 months 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

4 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 months 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

4 months ago

0 Follower 6 Views

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.

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