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+1 vote
18.3k views
in Physics by (75.1k points)
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In the circuit shown in fig, the reading of ammeter is....

(A) 1A 

(B) 2A 

(C) 3A 

(D) 4A

1 Answer

+2 votes
by (70.5k points)
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Best answer

The correct option (A) 1A

Explanation:

This can be reduced as 1Ω series with 1Ω hence R = 2Ω

This 2Ω is parallel with 2Ω hence R' = (1/1)Ω

Now R' in series with 2Ω hence R" = 1 + 2 = 3Ω

 R" parallel with 6Ω hence R"' = 3Ω || 6Ω

= 2Ω

This R"' & 2Ω in series. hence effective resistance is 2Ω + 2Ω = 4Ω

∴ Total current = I = {(12V)/(4Ω)}

∴ I = 3A

The circuit can be drawn as

 

at point C, I will get divide as (2I / 3) & (I/3).

further at point D, (2I/3) will get divided equally as (I/3), (I/3)

∴ current through ammeter = (I/3) = (3/3) = 1A 

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