Question

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?​

Answer 1

Correct 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:

• The current (I) drawn is 4 A

Given:

1. Current in circuit a (I₁) = 6 A.
2. Let Current in circuit b be I₂

Explanation:

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Now lets consider figure 'a'

As the three bulbs each of resistance R are connected in parallel, then their equivalent resistance will be,

$\large\bigstar\;\underline{\boxed{\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}}}$

Substituting the values,

$\longrightarrow\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}\\\\\\\longrightarrow\sf \dfrac{1}{R_{eq}}=\dfrac{1+1+1}{R}\\\\\\\longrightarrow\sf \dfrac{1}{R_{eq}}=\dfrac{3}{R}\\\\\\\longrightarrow\sf R_{eq}=\dfrac{R}{3}\\\\\\\longrightarrow{\boxed{\sf R_{eq}=\dfrac{R}{3}\;\Omega}}$

Hence, equivalent resistance is R/3 Ω.

Now, lets find Voltage in the circuit 'a'

From Ohm's law we know

$\large\bigstar\;\underline{\boxed{\sf V=I\;R}}$

Substituting the values,

$\longrightarrow\sf V = 6\times \dfrac{R}{3}\\\\\\\longrightarrow\sf V=2R\\\\\\\longrightarrow\boxed{\sf V=2R\; volts}$

∴ We got the potential as 2 volts.

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Now, this potential is same in circuit 'b'. and we need not consider fused resistor R.

Now, lets find equivalent resistance of rest two bulbs.

$\large\bigstar\;\underline{\boxed{\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}}}$

Substituting the values,

$\longrightarrow\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R}\\\\\\\longrightarrow\sf \dfrac{1}{R_{eq}}=\dfrac{1+1}{R}\\\\\\\longrightarrow\sf \dfrac{1}{R_{eq}}=\dfrac{2}{R}\\\\\\\longrightarrow\sf R_{eq}=\dfrac{R}{2}\\\\\\\longrightarrow{\boxed{\sf R_{eq}=\dfrac{R}{2}\;\Omega}}$

Hence, in circuit b equivalent resistance is R/2 Ω.

Now, Ohm's law we know,

$\large\bigstar\;\underline{\boxed{\sf V=I\;R}}$

Substituting the values,

$\longrightarrow\sf 2R = I\times \dfrac{R}{2}\\\\\\\longrightarrow\sf I=2R\times \dfrac{2}{R}\\\\\\\longrightarrow\sf I=2\times 2\\\\\\\longrightarrow\sf I=4\\\\\\\longrightarrow \large{\underline{\boxed{\textsf{\textbf{I = 4\;A}}}}}$

∴ The current (I) drawn from battery is 4 A.

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