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New answer posted

5 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

5 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

↑? (1) ε = 3
(2) ε – Ir = 2.5 V
⇒ Ir = 0.5
Now, IR = 2.5
⇒ R/r = 5.
⇒ PR/Pr = I²R/I²r = R/r = 5
⇒ Pr = 0.5/5 = 0.1

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Parallel of  10 k Ω and 400 Ω = 384.61 Ω

V 400 Ω = 384.61 384.61 + 800 * 6 = 1.95 V

New answer posted

5 months ago

0 Follower 2 Views

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

5 months ago

0 Follower 14 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

5 months ago

0 Follower 4 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.

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