Question

$\huge\underline\bold\red{Question:-}$
Three identical bulbs are connected in parallel with battery the current drawn from the battery is 6A . If one of the bulb gets fuse what will be the total current drawn from the battery?​

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Answer 1

Answer:

Initially, for a parallel connection

1/Req = 1/R + 1/R + 1/R (where R is the resistance of each bulb)

Req = R/3

I = 6A

Hence V = IReq

V = 6*R/3 = 2R

If one of the bulb is fused, then the current will still flow in the remaining 2 bulbs. The voltage will remain the same as the battery is same but the current will change.

The new equivalent resistance would then be

1/Req = 1/R + 1/R i.e. Req = R/2

Hence I = V/Req = 2R/R/2 = 4 A

hope this help

Answer 2

$\large\underline\color{grey}{\sf{Solution\: : - }}$

$\:\:\:\:\:\:\:\:\:\:$ Let , the potential difference maintained by the battery be $\color{red}{v_1}$ and let the resistance of each bulb be $\color{red}{R}$.

The equivalent Resistance of the circuit $\color{red}\dfrac{1}{r}$,

$\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$ $\dfrac{1}{r}$ + $\dfrac{1}{r}$ + $\dfrac{1}{r}$

$\:\:\:\:\:\:\:\:\:\:$ or , $\hookrightarrow$$\color{black}\boxed{ \dfrac{R}{3}}$

The current is ,

$\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$ $\color{black}\boxed{ i \: =\dfrac{v}{r} \: = \: \dfrac{3v}{R}}$

It is given that this current is $\color{red}{6A}$ .

So,

$\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$ $\dfrac{V}{R}$ = 2A

$\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$ $\tt\: 6A = \dfrac{3v}{R}$

If one of the bulbs gets fused only two bulbs remain connected in parallel.The equivalent Resistance $\color{red}{r'}$in that case is given by

$\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$ $\dfrac{1}{r'}$ = $\dfrac{1}{R}$ + $\dfrac{1}{R}$

$\:\:\:\:\:\:\:\:\:\:$or, $\color{black}\boxed{ r'\: =\: \dfrac{R}{2}}$

The current in the battery will be

$\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$ $\bold {i' }$ = $\dfrac{V}{R'}$ = $\dfrac{0v}{R'}$ = 2 × (2A)

$\:\:\:\:\:\:\:\:\:\:$$\star\;{\underline{\boxed{\pmb{\green{\frak{\;Answer\; : -\:4A}}}}}}$

$\large\underline\color{green}{\sf{Alternative\: : - }}$

$\:\:\:\:\:\:\:\:\:\:$ As three bulbs are identical they will draw equal currents .As the total current is 6A, each bulb will be drawn 2A of current when one bulb gets fused , there is no current through it .Each of the remaining bulbs keep connected to the battery as before. So the current through each is still 12A giving the total current 4A through battery.

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Hence , The correct Answer is 4A.

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