Suppose a vacuum cleaner uses 120 J of electrical energy. If 45 J are used to pull air into the vacuum cleaner, how efficient is
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Answer:
1) current = I
2) Resistance = V/I
3) current = 2I
4) resistance = V/2I
5) current = 3I
6) Resistance = V/3I
7) Current = 4I
8) Resistance = V/4I
Explanation:
When one bulb is connected across the battery then let say the current is given as I
Then resistance is given as
When two bulbs are in parallel with the battery then
total current becomes twice of initial current
so we have
current = 2I
Resistance of the circuit is now
When three bulbs are in parallel with the battery then
total current becomes three times of initial current
so we have
current = 3I
Resistance of the circuit is now
When four bulbs are in parallel with the battery then
total current becomes four times of initial current
so we have
current = 4I
Resistance of the circuit is now