Solve RC Circuits: Instantaneous Voltage in Capacitors

AI Thread Summary
To find the instantaneous voltage in a capacitor within an RC circuit, the key equation is Vc = E(1 - e^(-t/T)), where T is the time constant (RC). The negative exponent indicates the exponential decay of voltage as the capacitor charges towards the input voltage E. The "1-" term reflects that the capacitor voltage approaches E over time during charging, while during discharging, the equation changes. Understanding the relationship between current and voltage through differentiation and integration is crucial for solving these circuits. This foundational knowledge is essential for mastering RC circuits in electronics.
abk80
Messages
2
Reaction score
0
[SOLVED] Help with RC circuits

Can someone explain how to find the intantaneous value of voltage in a capacitor at any specific instant in time. I am taking a course in electronics and the textbook isn't really clear on this. I have the equation:
-t/T
Vc=E(1-e )
t=RC
Not sure why I am subracting from one. And why the negative symbols. I am sure this question is nonsense to an engineer or tech but I am stuck. Thanks
 
Engineering news on Phys.org
The fundamental equation to use when calculating the voltages and currents for a capacitor is the following:

i(t) = C\frac{dv(t)}{dt}

If you know the v(t) for the capacitor, you differentiate to get the i(t). If you know the i(t) for the capacitor, you integrate to get the v(t).

If you have a step change in current, you can solve the differential equation assuming a solution of the form:

v(t) = Ve^{\frac{-t}{\tau}} subject to initial conditions, and where tau is related to the R and C values in the circuit.

So when you solve this differential equation for a series RC circuit where there is a step change in voltage across the whole RC, you end up with a solution for the voltage across the capacitor that looks something like:

v(t) = V_i * (1-e^{\frac{-t}{RC}})

You get the "1-" term, because the capacitor voltage exponentially approaches the full input voltage. If instead you were discharging the capacitor through a resistor, then you don't get the "1-" term.

This page at wikipedia.org may be of help to you too: http://en.wikipedia.org/wiki/RC_circuit

Welcome to the PF!
 
Last edited:
thank you that really helped.
 
Hi all I have some confusion about piezoelectrical sensors combination. If i have three acoustic piezoelectrical sensors (with same receive sensitivity in dB ref V/1uPa) placed at specific distance, these sensors receive acoustic signal from a sound source placed at far field distance (Plane Wave) and from broadside. I receive output of these sensors through individual preamplifiers, add them through hardware like summer circuit adder or in software after digitization and in this way got an...
I have recently moved into a new (rather ancient) house and had a few trips of my Residual Current breaker. I dug out my old Socket tester which tell me the three pins are correct. But then the Red warning light tells me my socket(s) fail the loop test. I never had this before but my last house had an overhead supply with no Earth from the company. The tester said "get this checked" and the man said the (high but not ridiculous) earth resistance was acceptable. I stuck a new copper earth...
I am not an electrical engineering student, but a lowly apprentice electrician. I learn both on the job and also take classes for my apprenticeship. I recently wired my first transformer and I understand that the neutral and ground are bonded together in the transformer or in the service. What I don't understand is, if the neutral is a current carrying conductor, which is then bonded to the ground conductor, why does current only flow back to its source and not on the ground path...
Back
Top