I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW! A 12v Light bulb uses 24v of power.

a) Calculate its resistance and the current at full power.

b) Calculate the resistance of a 100w mains light bulb and the current at full power.

c) Two bulbs are connected in series and supplied from the mains power at 230v. Calculate the current in the circuit. Assume that both bulbs operate at their working resistance.

d) Use your answer to part c) to comment on the brightness of the bulbs.

a)

V = 12 VP = 24 W

=> R= 6 Ohms(Ω)

b)

Power (P)= 100 W

these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,

V = 240 V

R and 100 can interchange places

=> R = 576 Ω

By Ohm's Law:

=> 240 = I × 576

=>

=> I = 0.417 A

c)

I don't know it's resistance,... so sorry

d)

The brightness of the bulb in series is less than when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series is more than the individual resistance.

And

Brightness, i. e., Power

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

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