# Solving an RC circuit using superposition

Discussion in 'Homework Help' started by ecy5maa, Jan 1, 2011.

1. ### ecy5maa Thread Starter New Member

Nov 22, 2010
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Hello,

The circuit attached below is from Richard Dorfs Intro to Electric Circuits book. The switch is shut for a long time and opened at t=0.

The question asks to find the current i(t) across R11, which they use nodal analysis and find it to be 66.67 - 16.666 e micro Amps.

However when i do it from superposition 66.67 micro Amps is due to the 8 v source and + 16.666 e source due to the capacitor

My question is that if we use superposition instead, shouldnt the current due to the 8 v source and the capacitor be added since both send current towards R11 in the same direction?

Thus basicall I would just like to know why the current due to the capacitor will be subtracted?

Thank you in advance

P.S Unfortunately i dont have the book in question, although the answer is correct.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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In this circuit when you open the switch voltage across capacitor rise from 3.3V to 4V.
So capacitor will get extra charge, and this mean that in this circuit capacitor is charging when the switch is open.

3. ### ecy5maa Thread Starter New Member

Nov 22, 2010
6
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Yup that i got.

But what i dont get is that how come the current through R11 delivered by the capacitor is subtracted from the current delivered by the 8v source when the direction of current of both capacitor and 8v source is the same.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Hmm, but in this circuit capacitor do not deliver any Current to R11 .
When the switch is open R10 provide extra current to charge the capacitor.

5. ### ecy5maa Thread Starter New Member

Nov 22, 2010
6
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Yes, but since the switch is shut for a long time, and it is opened at t=0, there is already initial voltage across the capacitor...so shouldnt it be supplying current, till the current becomes zero across the capacitor..

Also basically i wanted to know if this was correct

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Capacitors supply current only when they are discharging.
In your circuit at t=0 voltage rise so there in no way to discharge the capacitor.
At T= 0 capacitor will be charge trough R10. So capacitor will be "absorbing" the current.

Last edited: Jan 1, 2011
7. ### ecy5maa Thread Starter New Member

Nov 22, 2010
6
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But wouldnt the capacitor discharge when the switch is opened?

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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No, capacitor star charging it self, because when the switch is open voltage wants to rise from 3.3V to 4V

9. ### ecy5maa Thread Starter New Member

Nov 22, 2010
6
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voltage wants to rise, but doesnt that mean that dv/dt != 0 hence there will be a voltage across the capacitor and hence a current that it supplies?

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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When the voltage on the capacitor rise, capacitor is charging ("absorbing" the current).
When the voltage on capacitor "drops", capacitor is discharging (supply the current).

And in your circuit capacitor is charging when switch is open and discharging when switch is close.

11. ### ecy5maa Thread Starter New Member

Nov 22, 2010
6
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ok thanks...but can you just clear one thing.....since Vo across the capacitor is not zero...should there or shouldnt there be current across the capacitor when the switch is opened.

I dont mean to say that the capacitor will be supplying current, but that there will be current across the capacitor...given by Io e^(-t/RC)?

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,175
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Yes the current will be flow through capacitor because voltage wants to rise, the extra charge is needed on capacitor plates. So the need of this extra charge on capacitor plates cause that current will be flow into capacitor plates (absorbing the current).

Last edited: Jan 1, 2011