Whatever energy the electrons have when they come out of one terminal of the battery, they completely use it up on their way around the circuit, and they stagger back into the other terminal of the battery totally exhausted, with no energy left.
If each coulomb of electrons has 6 joules of energy when they leave the battery, then that's the energy they'll give up to the circuit before they return to the battery.
For each coulomb of charge that moves through that circuit, each flashlight bulb
will take 3 joules of energy away from that coulomb, and turn the energy into heat
and light.
==> 2 bulbs, 3 joules per coulomb that flows through each bulb, total 6 joules
per coulomb that flows around the circuit.
Note:
The question says that the bulbs are in series, but that wasn't necessary.
The energy consumed by the bulbs would be the same if they're in parallel.
A cool extra factoid:
The battery gives each coulomb of electrons that leaves it 6 joules of energy.
There's a special name for "1 joule per coulomb of charge". That's the "<em>volt</em>".
A battery that gives each coulomb of charge 6 joules of energy is a 6-volt battery.
Answer:
Formula : =IF(E5>=4, "Y", "N")
Formula Explanation : =IF(cell_name>=4, [value_if_true], [value_if_false]
Explanation:
After Writing this Formula in cell H5 Drag it to the cell till where you want.
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