Statement: Two light bulbs having the watt rating of 90 watts and 45 watts are connected in series with a 230-volt source (wall socket). Which bulb will glow brighter?

Solution:

We can calculate the resistance by using the Ohm's law.

From Ohm's law: R = V

I = V/R = 230/1762 = 0.130 A

In present case, the actual power consumed by the lamps can be calculated by the equation, P = I

For 90 > P = (0.130)

For 45 > P = (0.130)

*Power is rated for 230-volt input.*Solution:

We can calculate the resistance by using the Ohm's law.

From Ohm's law: R = V

^{2}/P**For first bulb: R = (230)**^{2}/90 watts = 587 Ω

**For first bulb: R = (230)**^{2}/45 watts = 1175 Ω**The overall current flowing through the circuit will be:**

I = V/R = 230/1762 = 0.130 A

In present case, the actual power consumed by the lamps can be calculated by the equation, P = I

^{2}RFor 90 > P = (0.130)

^{2}* 587 Ω = 10 wattFor 45 > P = (0.130)

^{2}* 1175 Ω = 19.85 watt
result of this question

ReplyDeleteThese bulbs are energy skillful, don't incorporate risky components like mercury. They don't emit heat like radiant bulbs. motion sensor bulbs

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