Question

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

Answer 1

Answer :-

If one bulb gets fused, then current drawn from the battery is 4 A .

Explanation :-

Since the three bulbs are identical, so their resistance will same. When three bulbs are connected in parallel, then the equivalent resistance (Rₑ) :-

⇒ 1/Rₑ = 1/R + 1/R + 1/R

⇒ 1/Rₑ = 3/R

⇒ R = 3Rₑ

⇒ Rₑ = R/3

Now potential difference supplied by the battery, according to Ohm's Law will be :-

V = I × Rₑ

⇒ V = 6 × R/3

⇒ V = 2R

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When one of the bulb gets fused, then new equivalent resistance of the remaining two bulbs (Rₑ') :-

⇒ 1/Rₑ' = 1/R + 1/R

⇒ 1/Rₑ' = 2/R

⇒ R = 2Rₑ'

⇒ Rₑ' = R/2

As the bulbs were connected parallely, so potential difference will be same as earlier. Thus, let's calculate the current drawn using Ohm's Law .

V = I' × Rₑ'

⇒ 2R = I' × R/2

⇒ I' = 2R/(R/2)

⇒ I' = 4R/R

⇒ I' = 4 A

Answer 2

Question:

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• Three identical bulbs are connected in parallel with battery the current drawn from the battery is 6 A. If one of the bulb gets fuse what will be the total current drawn from the battery?

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Answer:

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• Total current drawn from the battery after one bulb gets fuse is 4 A.

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ㅤㅤㅤStep - By - Step Explanation

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$\underline{\pmb{\bf{\bigstar\:\purple{Given\:that}\:.\:.\:.}}}$

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• Current in circuit = 6 A

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$\underline{\pmb{\bf{\bigstar\:\purple{To\:Find}\:.\:.\:.}}}$

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• Total current drawn from battery if one bulb gets fuse?

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$\underline{\pmb{\bf{\bigstar\:\purple{Required\:Solution}\:.\:.\:.}}}$

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• Let resistance of three bulbs be R, we know that they are connected in parallel. So, their equivalent resistance will be ::

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$\bigstar\:\underline{\boxed{\pmb{\bf{\pink{\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}}}}}}}$

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$\underline{\pmb{\bm{\bigstar\:Putting\:all\:values\:.\: . \:.}}}$

$\\ \qquad:\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} \\ \\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1 + 1 + 1}{R} \\ \\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{3}{R} \\ \\ :\implies\:\underline{\boxed{\pmb{\mathscr{\red{R_{eq} = \dfrac{R}{3}\:\Omega}}}}}\:\bigstar$

$\\ \therefore\:{\underline{\sf{Hence,\:equivalent\:resistance\:is\:\pmb{\frac{R}{3}\:\Omega}}}}$

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• Now let's find voltage in the circuit by using ohm's law ::

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$\bigstar\:\underline{\boxed{\pmb{\bf{\pink{V = IR}}}}}$

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$\underline{\pmb{\bm{\bigstar\:Putting\:all\:values\:.\: . \:.}}}$

$\\ :\implies\:\sf V = \cancel{6}\:\times\:\dfrac{R}{\cancel{3}}$

$\\ :\implies\:\sf V = 2\:\times\:R$

$\\ :\implies\:\underline{\boxed{\pmb{\mathscr{\red{V = 2R\:volts}}}}}\:\bigstar$

$\\ \therefore\:{\underline{\sf{Hence,\:potential\:is\:\pmb{2R\:volts}}}}$

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• Equivalent resistance after one bulb gets fuse ::

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$\bigstar\:\underline{\boxed{\pmb{\bf{\pink{\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}}}}}}$

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$\underline{\pmb{\bm{\bigstar\:Putting\:all\:values\:.\: . \:.}}}$

$\\ \qquad:\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R} \\ \\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1 + 1}{R} \\ \\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{2}{R} \\ \\ :\implies\:\underline{\boxed{\pmb{\mathscr{\red{R_{eq} = \dfrac{R}{2}\:\Omega}}}}}\:\bigstar$

$\\ \therefore\:{\underline{\sf{Hence,\:equivalent\:resistance\:is\:\pmb{\frac{R}{2}\:\Omega}}}}$

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• Now, let's find the current drawn from battery by using ohm's law ::

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$\bigstar\:\underline{\boxed{\pmb{\bf{\pink{V = IR}}}}}$

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$\underline{\pmb{\bm{\bigstar\:Putting\:all\:values\:. \:. \:.}}}$

$\\ :\implies\:\sf 2R = I\:\times\:\dfrac{R}{2}$

$\\ :\implies\:\sf I = 2\cancel{R}\:\times\:\dfrac{2}{\cancel{R}}$

$\\ :\implies\:\sf I = 2\:\times\:2$

$\\ :\implies\:\underline{\boxed{\pmb{\mathscr{\red{I = 4\:A}}}}}\:\bigstar$

$\\ \therefore\:{\underline{\sf{Hence,\:current\:drawn\:from\:battery\:is\:\pmb{4\:A}}}}$

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