Question

$\huge\boxed\bf{Question}$

B1 B2, and B3 three identical bulbs connected as shown in the figure. When all the three bulbs glow, a current of 3 A is recorded by the ammeter A.
(a) What happens to the glow of the other two bulbs when the bulb Bi gets fused?
(b) What happens to the reading of A1,A2, A3 and A when the buib Bz gets fused?​

Answer 1

Answer:

1. the glow of the bulbs B2 and B3 will remain the same as the V and R remains the same.

2. When bulb 2 gets fused:

Reading on A1 is 1 ampere, reading on A2 is 0 ampere, reading on A3 1 ampere and reading on A is 2 ampere.

3. Assuming the voltage being V the resistance of the circuit is:

R=V/I=V/3

I hope this may help you♥️

Answer 2

$1)the \: glow \: of \: the \: bulb \: b_{2} \: and \: b_{3} \: will \: remain \: same \: because \: glow \: of \: bulbs \: depend \: on \: power. \\ \\ power = \frac{ {v}^{2} }{r} \\ since \: potential \: difference \: of \: v \: and \: resistent \: r \: of \: b_{2} \: and \: b_{3} remain \: same \: so \: glow \: of \: the \: b_{2} \: and \: b_{3} \: will \: also \: remain \: same.$

$2)in \: parallel \: connections \: \\ \frac{1}{r} = \frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3 } = \frac{3}{r} \: \: \: \: (as \: bulbs \: are \: identical \: so \: r1 = r2 = r3 = r) \\ r = \frac{r}{3} \\ v = ir \\ 4.5 = 3 \times \frac{r}{3} \\ r = 4.5 \\ but \: as \: b2 \: gets \: fused \: so \: only \: 2 \: bulbs \: are \: connected \: in \: parallel \: \\ therefore \ \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r} = \frac{2}{4.5} \\ r = \frac{4.5}{2} \\ i = \frac{v}{r} = \frac{2 \times 4.5}{4.5} = 2a \\ since \: b2 \: gets \: fused \: and \: only \: b1 \: and \: b3 \: are \: in \: parallel \: and \: have \: same \: resistent \: so \: 2a \: will \: be \: divided \: equally \: between \: them \: so \: 1a \: to \: each$

$3)p = v \times 1 = 4.5 \times 3 = 13.5w$

Hope it helps