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New answer posted

20 hours ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

20 hours ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

It is balanced wheatstone bridge

 

New answer posted

20 hours ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Charge remains same

 

New answer posted

2 days ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

2 days ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

It is balanced wheatstone bridge

 

New answer posted

2 days ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 days ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Since Gand S are in parallel

-> VG = VS

->3mA * 10 = 8A * RS

->RS = 3.75mW

New answer posted

4 days ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

I . d t = 0 1 0 ( 2 0 + 3 t ) d t

= ( 2 0 t ) 0 1 0 + 3 ( t 2 2 ) 0 1 0 = 3 5 0 C

New answer posted

4 days ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Equivalent circuit

i = 9 4 + 1 = 1 . 8 ? A

i 2 = 0 . 9 ? A

(VP– VQ) = 0.9 * 6 – 0.9 * 2

VC = 3.6V

U =  1 2 CV2 =   1 2 * 4 * 3.6 * 3.6mJ

= 25.92mJ

New answer posted

a week ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Current in circuit (I) = 8 2 6  = 1 A.

So, VC – 4 (1) – 2 + 8 – 2 (1) = VA

VC – 6 – 2 + 8 = VA

VC – VA = 0 V

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