(20 MARKS)Urgent!!! About Electricity!!!! PLEASE HELP!!

睇power,power高d就光d

如果power一樣
the one with higher resistance will be brighter
V=IR
as in the same circuit, the current is the same
by formula, V is greater in larger R
therefore the one with higher resister get higher voltage and hence more energy(light)

the current will decrease as it pass the circuit
because the battery is losing energy(V is decreasing)
by V=IR, current drops
thats why u see the bulb become dimmer and dimmer

The one with lower wattage will be brighter, when 2 bulbs are connected in series (assuming both bulbs are rated at the same voltage).

The reason is simple. Since power (wattage) for a bulb is inversely proportional to the square of its resistance, this implies that a lower wattage bulb has a higher resistance when compared to one having a higher wattage.

Now, when the 2 bulbs are connected in series, the current through both of them is the same. Therefore, the lower wattage bulb (which has a higher resistance) will dissipate more power.

Now, you can tell which one has a higher resistance by comparing the brightness of the bulbs.

The last part. A series circuit means the total resistance is higher. Hence, for the same applied voltage, the current will be less.

The bulb which consumes higher power will be brighter. This is because the higher power it consumes, the more light energy it will give.

From Power = (Current)² * (Resistance). Since they are connected in series, the current flowing through the 2 light bulbs is the same. From the equation, it is clear that the bulb with higher resistance consumes more power and hence it will be brighter.

Will the current be decreased as it goes around the series circuit?
No, it won't. This is because the current flowing through each component is the same in a series circuit. Instead, their potential differences are difference because of their unequal resistance. It follows from Ohm's law which can be expressed as Voltage = Current * Resistance.