What happens to the reading of $A_{1}, A_{2}, A_{3}$ and A when the bulb $B_{2}$ gets fused?

A_{1} = 1 A, A_{2} = 0 A, A_{3} = 0 A

and

A = 2 A

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A_{1} = 1 A, A_{2} = 0 A, A_{3} = 1 A

and

A = 2 A

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A_{1} = 2 A, A_{2} = 0 A, A_{3} = 1 A

and

A = 2 A

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A_{1} = 1 A, A_{2} = 0 A, A_{3} = 1 A

and

A = 1 A

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Solution

The correct option is B $A_{1} = 1 A, A_{2} = 0 A, A_{3} = 1 A$ and $A = 2 A$
In parallel connections of the circuit, the total resistance across is calculated as follows.
$\frac{1}{R^{^{'}}} = \frac{1}{R 1} + \frac{1}{R 2} + \frac{1}{R 3} = \frac{3}{R}$ , As bulbs are indentical and so R1 = R2 = R3 = R
Therefore, R' = R/3.
According to Ohm's law, V = IR.
In this case, the voltage is given as 4.5 V and the current across the resistances is 3 A.
Substituting them in the above formula we get, $4.5 = 3 \times (R / 3)$
Hence, the resistance R1=R2=R3=R = 4.5 ohms
But as the bulb B2 gets fused only two bulbs are connected in parallel.
Therefore, $\frac{1}{R^{^{'}}} = \frac{1}{R 1} + \frac{1}{R 2} = \frac{3}{4.5}, t h a t i s, R^{^{'}} = \frac{4.5}{2} o h m s .$
So, $I = \frac{V}{R^{^{'}}} = \frac{2 \times 4.5}{4.5} = 2 A$ .
Since B2 gets fused and only B1 and B3 are in parallel and have same resistance so 2A current will be equally distributed between them as 1A to each.
Hence, the ammeter readings would be where A1 shows 1 ampere, A2 shows zero as bulb B2 blows off, A3 shows 1 ampere and A shows 2 amperes, the total current in the circuit.

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What happens to the reading of A1,A2,A3 and A when the bulb B2 gets fused?

What happens to the reading of $A_{1}, A_{2}, A_{3}$ and A when the bulb $B_{2}$ gets fused?