Question

8. study the circuit shown in which three identical bulbs b1, b2 and b3 are connected in parallel with a battery of 4.5 v. (1+2+2) (i) what will happen to the glow of other two bulbs if the bulb b3 gets fused? (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? (iii) find the total resistance of the circuit

Answer 1

Answer:

Step-by-step explanation:

(i) the bulb will glow neither brighter nor dull

(ii) add all the wattage (wattage=work=joule) you will get 4.5.

   now, divide it by the voltage given :

=> 4.5/4.5 = 1 A       (A = ampere = current = I)

(iii)now we have got total current = 1.5 A and voltage = 4.5 v

=> (ohms law) V = IR . R = V/I and R1 +  R2 + R3

=> R1 = 4.5/1.5 = 3 ; R1 = 3    

since, R1 = R2 = R3 = 3Ω : (all bulbs are identical )

Answer 2

Answer:

Below are the answers to each question for this problem.

Step-by-step explanation:

(i) what will happen to the glow of the other two bulbs if the bulb b3 gets fused?

Ans - The lightbulb won't get much brighter or duller.

(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?

Ans - When you sum up all the watts, you get 4.5. Now, divide it by the voltage $\frac{4.5}{4.5}$ = 1 A (A stands for ampere)

(iii) find the total resistance of the circuit

Ans - we have total current = 1.5 A and voltage = 4.5 v

As per Ohms law, V = IR.

R = $\frac{V}{I}$ and R1 +  R2 + R3

R1 = $\frac{4.5}{1.5}$

R1 = 3

Since, R1 = R2 = R3

= 3Ω and all bulbs are identical

Thus, the strength of a direct current is inversely related to the resistance of the circuit and directly proportional to the potential difference.