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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Answers
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: