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  • B1, B2 and B3 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. (i) What happens to the glow of the other two bulbs when the bulb B1 gets fused? (ii) What happens to

B1, B2 and B3 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. 

(i) What happens to the glow of the other two bulbs when the bulb B1 gets fused? 

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

(iii) How much power is dissipated in the circuit when all the three bulbs glow together? 

Answers (1)

(i) Nothing happens to the glow of bulbs B2 and B3  when B1 gets fused because all the bulbs are connected parallel therefore the power output won’t get changed.

P=\frac{V^{2}}{r}

          In parallel connection voltage across the bulb does not change and resistance is independent of the connections in the circuit as it is an intrinsic property of wire based on its dimensions

(ii) Given all the three bulbs are identical therefore all the bulbs have same resistance R 

All the three bulbs are in Parallel combination therefore

\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\\ \frac{1}{R_{eq}}=\frac{3}{R}\\ R_{eq}=\frac{R}{3}
Current in the ammeter A is 3A

3A=\frac{4.5}{\frac{R}{3}}\\ R=\frac{4.5}{\frac{3}{3}}\\ R=4.5\Omega

Resistance of each individual bulb is 4.5?

So reading in A1, A2, A3 when the bulb A2 gets short circuited is:

Reading in A1    

A_{1}=\frac{4.5}{4.5}=1A

So reading in A1 when A2 is short circuited is 1A.

Reading in A2 is zero as it is short circuited

Reading in A3    

A_{3}=\frac{4.5}{4.5}=1A

So reading in A3 when A2 is short circuited is 1A.

(iii) Power dissipated when all the bulbs glow together: As they are in parallel combination, voltage across each bulb is same. Also, all of them are identical therefore, we can find the power consumption of one bulb and multiplying thrice will give us the total power.
Tripling will give you total power consumed

Power \;in\; one\; bulb=\frac{V^{2}}{R}=\frac{4.5^{2}}{4.5}=4.5W

Total power consumed=4.5×3=13.5 W

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infoexpert24

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