Soil.............................
Answer:
16.7 mF
Explanation:
The total capacitance of two capacitors connected in series is given by the formula:
in our problem, we have:
C1 = 45 mF is the capacitance of the first capacitor
C2 = 26 mF is the capacitance of the second capacitor
Substituting into the equation, we find:
Well, Godess, that's not a simple question, and it doesn't have
a simple answer.
When the switch is closed . . .
"Conventional current" flows out of the ' + ' of the battery, through R₁ ,
then through R₂ , then through R₃ . It piles up on the right-hand side of
the capacitor (C). It repels the ' + ' charges on the left side of 'C', and
those flow into the ' - ' side of the battery. So the flow of current through
this series circuit is completely clockwise, around toward the right.
That's the way the first experimenters pictured it, that's the way we still
handle it on paper, and that's the way our ammeters display it.
BUT . . .
About 100 years after we thought that we completely understand electricity,
we discovered that the little tiny things that really move through a wire, and
really carry the electric charge, are the electrons, and they carry NEGATIVE
charge. This turned our whole picture upside down.
But we never changed the picture ! We still do all of our work in terms of
'conventional current'. But the PHYSICAL current ... the actual motion of
charge in the wire ... is all exactly the other way around.
In your drawing ... When the switch is closed, electrons flow out of the
' - ' terminal on the bottom of the battery, and pile up on the left plate of
the 'C'. They repel electrons off of the right-side of 'C', and those then
flow through R₃ , then through R₂ , then through R₁ , and finally into the
' + ' terminal on top of the battery.
Those are the directions of 'conventional' current and 'physical' current
in all circuits.
In the circuit of YOUR picture that you attached, there's more to the story:
Battery current can't flow through a capacitor. Current flows only until
charges are piled up on the two sides of 'C' facing each other, and then
it stops.
Wait a few seconds after you close the switch in the picture, and there is
no longer any current in the loop.
To be very specific and technical about it . . .
-- The instant you close the switch, the current is
(battery voltage) / (R₁ + R₂ + R₃) amperes
but it immediately starts to decrease.
-- Every (C)/((R₁ + R₂ + R₃) seconds after that, the current is
e⁻¹ = about 36.8 %
less than it was that same amount of time ago.
Now, are you glad you asked ?
