# Flying capacitor circuit problem

## Homework Statement

I've attached the problem in the pdf, which includes the circuit diagram, but I'll copy the text of the problem here:

(a) Draw the equivalent circuit when the switches are in the position indicated in the diagram and
calculate the potential difference that develops across the capacitor in the limit when t  1/R1C.
(b) At some later time, the both S1 and S2 switched simultaneously. Draw the equivalent circuit
and calculate the potential difference across R2 the instant after this event.
In this type of curcuit is commonly found in DC-DC converters. In such a circuit configuration,
C is called a flying capacitor.

## Homework Equations

Kirchhoff's laws, Ohm's law

V=IR for resistor, V=Q/C for capacitor

## The Attempt at a Solution

a) the equivalent circuit is the voltage source with R1 and then C in series

using Kirchoff's loop law then,

V-IR1-Q/C=0

differentiating, and then integrating to solve for I, I get

I(t)=V/R1*e^(-t/R1C)

which matches the equation derived in my book for an RC circuit.

then, Vc=V(1-e^-t/R1C) where Vc is the voltage across the capacitor, which also matches the equation in my book.

Then, if t>>R1C, I said that Vc-->V

b) The equivalent circuit here with the switches flipped is the voltage source, then R2, then C all in series.

The problem I'm having arises here: If the C is fully charged where Vc=V when the switch is flipped, then no current should be able to flow to R2, and there shouldn't be any voltage drop.

This answer seems too trivial though, and seems to render this kind of circuit pointless. What am I missing here?

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gneill
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The problem I'm having arises here: If the C is fully charged where Vc=V when the switch is flipped, then no current should be able to flow to R2, and there shouldn't be any voltage drop.

This answer seems too trivial though, and seems to render this kind of circuit pointless. What am I missing here?
Check the polarity of the voltage on the capacitor and do a Kirchoff voltage sum around the loop.

ahhh i think i see now. so the loop law would be V(battery)-V(capacitor)-V(R2)=0, so since Vb=Vc, the voltage across the resistor is double the battery voltage. that makes a lot more sense. thanks!