Question

50 pts ✌️
ques in attachment .. ans plss
no faltu ans..
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Answer 1

Explanation:

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(i) There will be no effect on the glow of

the other two bulbs and will remain

same when B1 gets fused because

glowing of bulb depends on the power

and the potential difference and

resistance remains same of other two

bulbs.

_____________________

(ii) When there are parallel connections:

Net resistance will be

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3}$

Since resistance is same so,

R’ $= \frac{R}{3}$

Applying ohm’s law V= IR

R = 4.5Ω

Since, B2 gets fused, so now only two

bulbs B1 and B3 are in parallel.

∴ net resistance in parallel

1/R’ = 2/R

R’ = 4.5/2 Ω

$I = V/R = \frac{2×4.5}{4.5}$

$I = 2A$

So, current will be distributed in both

the bulbs as 1 A each.

_____________________

(iii) Power dissipated when all three

bulbs glow together.

P = V × I

P= 4.5 × 3 = 13.5 W

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Answer 2

Answer:

(i) There will be no effect on the glow of

the other two bulbs and will remain

same when B1 gets fused because

glowing of bulb depends on the power and the potential difference and resistance remains same of other two

bulbs.

(ii) When there are parallel connections:

Net resistance will be

Since resistance is same so,

R' = R

Applying ohm's law V= IR

R = 4.50

Since, B2 gets fused, so now only two

bulbs B1 and B3 are in parallel.

:: net resistance in parallel

1/R' = 2/R

R' = 4.5/2 Q

I=V/R= 2×4.5 4.5

I = 2A

So, current will be distributed in both

the bulbs as 1 A each.

Yill 63

(iii) Power dissipated when all three

bulbs glow together.

P= V XI

P= 4.5 x 3 = 13.5 W