An idea cell (internal resistance $= 0$ ) is connected to two identical bulbs as shown in figure, if the bulb $B_{2}$ get fused, the current in $B_{1}$

increases

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Decreases

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Remian same

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Mav increase or decrease

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Solution

The correct option is C Remian same
In parallel connection, potential difference is equal

So, current in each bulb, $I_{1} = \frac{V}{R}$

If one bulb is fused, potential still remain same.

So, current in bulb is, $I_{2} = \frac{V}{R}$

Hence, current remain same.

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Similar questions

B_{1}, B_{2}

and

B_{3}

are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
What happens to the glow of the other two bulbs when the bulbs

B_{1}

gets fused?

Question 28 (i)
$B_{1}, B_{2}$ and $B_{3}$ are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
(i) What happens to the glow of the other two bulbs when the bulb $B_{1}$ gets fused?

Question 28 (ii)
$B_{1}, B_{2}, a n d B_{3}$ are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
(ii) What happens to the reading of $A_{1}, A_{2} A_{3}$ and A when the bulb $B_{2}$ gets fused?