RC Circuits, how do you solve for V(infinity)?

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SUMMARY

The discussion focuses on solving for the voltage across a capacitor in an RC circuit as time approaches infinity (V(infinity)). The relevant equation provided is v(t) = v(infinity) + [v(t0)-v(infinity)]e^[-(t-t0)]. Participants clarify that at steady-state DC, the capacitor behaves as an open circuit, meaning V(infinity) equals the battery voltage (VT). The conversation emphasizes understanding the time-dependent behavior of the capacitor and the circuit's configuration before and after the switch is activated.

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sugz
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Homework Statement


Question is attached below

Homework Equations


v(t) =v(infinity) + [v(t0)-v(infinity)]e^[-(t-t0)]

The Attempt at a Solution


For t<0, you get the circuit where it is only Rt, R, and C, but in dc circuits, the capacitor is an open circuit, does this mean v(t0) =0?

For t>0 the middle Rt is removed and you get a circuit consisting of Vt, Rt, R and C. To find Rth, you short circuit Vt and open circuit C, so Req= Rt+R. So time constant is C(Rt+R). To find v(infinity), we open circuit the the capacitator, but how do you solve for v(infinity)?
 

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How large a voltage do you need across the capacitor in order that the current from the battery to the capacitor is zero?
 
I'm not too sure to be honest!
 
Hi,

So far, what I know is that the voltage across the capacitor is zero when t<0. For t>0,the switch is in position b and the capacitor is an open open circuit do dc. This will result in equation:

-VT+(RT+R)*ic+Vc(t>0)=0
= Vc(t>0)=VT-(RT+R)*ic

Can someone help me please?
 
"At DC a capacitor appears as an open-circuit" holds true ONLY under steady DC when all currents and voltages have reached fixed, unchanging levels. You could say this occurs at t=infinity.

At switch-on when the capacitor voltage is not at its final fixed value, the capacitor for that brief instant appears as a short-circuit.
 
1. Homework Statement
Question is attached below
Because print in photos is ALWAYS more difficult to read than forum text, it would be appreciated if you would ALWAYS retype the text in photos into your post out of consideration for those forum participants contending with small screens or thick lenses, or both. :smile:
 
sugz said:
I'm not too sure to be honest!
Hint: When the voltage across the capacitor (VC) is equal to the battery voltage (VT), what is the current from the battery to the capacitor?
 
Dear sugz,

In post #1 you mention one relevant equation. In #4 another one, but without explicitly showing where the time dependencies are.
I am wondering if you know what the one in post #1 means and understand where it comes from.
If you don't feel all that comfortable in this exercise, I'd advise you to read up just a little on capacitors, here and here and here. Or in your textbook, if that is clear enough.
The subject is not all that complicated and well worth the investment of your time.
I could repeat the material, but find the hyperphysics hard to surpass in clarity.
 
sugz said:
So far, what I know is that the voltage across the capacitor is zero when t<0. For t>0,the switch is in position b and the capacitor is an open open circuit do dc.

Correct but...

a) What is the voltage on the capacitor at t=0 ?
b) What is the voltage drop across the resistors at t=0?
c) What is the current through the resistors at t=0?
d) What happens to the charge and voltage on the capacitor as a result of that current?
 

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