I need help on ideas for a science project.. (highschool ideas)
1 answer:
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The equivalent resistance of several devices connected in parallel is given by
where
are the resistances of the various devices. We can see that every time we add a new device in parallel, the term 
increases, therefore the equivalent resistance of the circuit 
decreases.
But Ohm's law:
tells us that if the equivalent resistance decreases, the total current in the circuit increases. The power dissipated through the circuit (and so, the heat produced) depends on the square of the current:
therefore if there are too many devices connected in parallel, this can be a problem because there could be too much power dissipated (and too much heat) through the circuit.
where
But Ohm's law:
tells us that if the equivalent resistance decreases, the total current in the circuit increases. The power dissipated through the circuit (and so, the heat produced) depends on the square of the current:
therefore if there are too many devices connected in parallel, this can be a problem because there could be too much power dissipated (and too much heat) through the circuit.
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. , the amount of charge stored in this capacitor, will stay the same.
The formula relates the electric potential across a capacitor to:
, the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential
by changing the capacitance
of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of
will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of
becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.