Question

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Three identical bulbs are connected in parallel with battery the current drawn from the battery is 6A . If one of the body gets future what will be the total current drawn from the battery?​
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Answer 1

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Three identical bulbs are connected in parallel with battery. The current drawn from the battery is 6A. If one of the body gets fused, what will be the total current drawn from the battery?

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• Three identical bulbs are connected in parallel with battery.

• The current drawn from the battery is 6A .

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Three identical bulbs are connected in parallel with battery.

The current drawn from the battery is 6 A.

So, 3 bulbs uses 6 A of current.

➳ 1 bulb uses $(\frac{6}{3})$A = 2 A.

As 1 bulb out of 3 gets fused, the fused bulb will not used electricity.

So, the current will be drawn by 2 bulbs.

➳ 1 bulb uses 2 A of current.

➳ 2 bulbs use = (2 × 2) A = 4 A.

So, 2 bulbs use 4 A of current.

As a result, 4 A of current will be drawn from the battery.

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4 A of current will be drawn from the battery after 1 bulb gets fused.

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Answer 2

Answer:

• Total current drawn from battery when one of the three bulbs will get fused = 4 Amperes.

Explanation:

Given that, Three identical bulbs are connected in parallel combination

so Let, individual resistance of each bulb be r

then, Equivalent resistance of three bulbs would be

→ 1/R(eq) = 1/r  +  1/r  +  1/r

→ 1/R(eq) = 3/r

→ R(eq) = r/3

Now, given that current drawn from the battery is

I = 6 Amperes

so, Using ohm's law expression

→ V = I · R(eq)

→ V = ( 6 ) · ( r/3 )

→ V = 2 r  

Now, given that

One of the bulb gets fused so,

New equivalent resistance of combination would be

→ 1/R'(eq) = 1/r  +  1/r

→ 1/R'(eq) = 2/r

→ R'(eq) = r/2

In this condition,

Voltage will remain the same, but Current will change because now only two bulbs will draw the current

so,

Again using expression for ohm's law

→ V = I' · R

→ 2 R = ( I' ) · ( r/2 )

→ I' = 4   Amperes

Therefore,

• Total current drawn from the battery when one of the bulb would be fused is 4 Amperes.