Question

A bulb of resistance 529 ohm works on 230 v . the filament was joined together and made to work . let the resistance of the filament now be 460 ohm . what is the change in the power of the bulb ? what is the current flowing through the bulb?


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Answer 1

Answer:

Part (a):

1. Given: R1 = 529 Ω, R2 = 460 Ω, V = 230 volts.

2. When the bulb was manufactured, it had a certain resistance. It was 529 Ω.

• So the manufacturer knows how much energy it will consume in one second (if 230 V is supplied) .

• We can calculate it ourselves using the equation which relates power to [voltage, resistance]  

• We can use Eq.10.5: P = V2⁄R

• Substituting the values, we get: P1 = 2302⁄529 = 100 W.

3. When the new resistance is 460, the new power = P2 = 2302⁄460 = 115 W.

So change in power = (115 - 100) = 15 W

4. The power has increased. That means, more energy is consumed, which is equivalent to saying: 'more energy is produced'.

• So more heat will be produced. But the filament may not be able to withstand this greater heat. Thus it will melt and break again  

5. Another explanation:

• We have seen in the previous section that, when the broken pieces of the filament are joined together, the current will increase.  

• According to Joule's law, when current increases, heat produced also increases

• But the filament may not be able to withstand this greater heat. Thus it will melt and break again  

Part (b):

1. We are required to find the currents.

We can use Eq.10.6: P = VI, which relates power to [Voltage, current]  

2. Voltage remains constant at 230 V

So we can write:

• P1 = V × I1 ⟹100 = 230 × I1 ⟹ I1 = 100⁄230 = 0.4348 A  

• P2 = V × I2 ⟹115 = 230 × I2 ⟹ I2 = 115⁄230 = 0.5 A

3. So change in current = (I2 - I1) = (0.5 - 0.4348) = 0.0652

Explanation: