Question

pls answer this question​

Answer 1

Answer:

Explanation:

Solution,

(i) What will happen to the glow of the other two bulbs if the bulb B3 gets fused?

Answer, (i)

Other bulbs will glow with the same brightness.

(ii) if the wattage of each bulb is 1.5 W how much reading will the ammeter A show when all the three bulbs glow simultaneously.

Answer, (ii)

When the bulbs are in parallel,

The wattage will be added as 4.5 W.

Then,

= 4.5 W/4.5 V

= 1 A.

(iii) Find the total resistance of the circuit.

Answer, (iii)

Since the ammeter rating is 1 A,

Then,

The resistance of the combination would be,

= 4.5 V/1 A

= 4.5 Ohm

Answer 2

Given

➺ B₁, B₂ and B₃ are in parallel with a potential difference of 4.5v

i) What will happen to the glow of other two bulbs if bulb B₃ gets fused.

☛ As we know that, in parallel combination if 2-3 bulbs are connected and one of them is fused, others won't get affected for that as they are in parallel.

☛ So to say, if one bulb gets fused other bulbs will remain as same as before, so they will glow as same as before!!

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ii) If the wattage of each bulb is 1.5 W, how much reading will the Ammeter A show when three bulbs glow simultaneously?

As wattage of 1 bulb is 1.5 W

➝ Wattage of 3 bulbs = (3 × 1.5) W

➝ Wattage of 3 bulbs = 4.5 W

Now we have potential difference (V) = 4.5 v

Now using formula :

☛ Power (P) = Potential difference(V) × Current (I)

➝ 4.5 = 4.5/I

➝ I = 4.5/4.5

➝ I = 1 A

∴ Ammeter will show 1 Amphere reading.

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iii) Now we have,

• Potential difference (V) = 4.5v
• Current (I) = 1 A
• Resistance (R) = ?

Now using Ohm's law :

☛ V = I * R

➝ 4.5 = 1 * R

➝ R = 4.5Ω

∴ Total resistance of circuit = 4.5Ω