Time constant of $RC$ circuit with more than one capacitor

homework-and-exercises electric-circuits charge electrical-resistance capacitance

2,091

You need to provide more details on the circuit. I'm going to assume you have two equal initially charged capacitors of capacitance $C$ in parallel which are then connected across a resistor $R$ by a switch initially in the open position. Of interest is then the equivalent time constant for the equivalent circuit that dictates the current in the circuit and the voltage across the resistor as a function of time after closing the switch.

The equivalent capacitance of capacitors in parallel is the sum of the capacitances, or in this case, $2C$. All of the energy stored in the equivalent capacitance is discharged through the single resistance $R$. Therefore the time constant for this circuit is R(2C) where $2C$ is the equivalent capacitance of the circuit.

The equation for the current in the circuit as a function after closing the switch at time t=0 for time equal to or greater than zero.

$$i(t)=\frac{v_{c}(0)}{R}e^{-\frac{t}{2RC}}$$

Where $v_{c}(0)$ = the initial voltage across the two capacitors prior to closing the switch. The voltage across the resistor is then simply $i(t)R$

you combined the two capacitors as one and then calculated the time constant..how should i proceed if i want to find the time constant of each individual capacitor.

the above comment is precisely my doubt

The time constant is a characteristic of the RC circuit, not a single circuit component (single capacitor). The following is from Wikipedia (italics are mine):

"The RC time constant, also called tau, the time constant (in seconds) of an RC circuit, is equal to the product of the circuit resistance (in ohms) and the circuit capacitance (in farads)..."

Notice it is the circuit capacitance, meaning in the case of your example the equivalent capacitance of the parallel combination. If you take one capacitor away, the time constant is simply halved. That means the time for the current (or voltage) to decay to 1/e ≈ 36.8% of the initial value would be half the time as two equal capacitors in parallel. This is simply because there is half the energy initially stored in the capacitance.

Hope this helps.

2,091

Dylan Rodrigues

Updated on July 26, 2022

Comments

Dylan Rodrigues 10 months

Suppose I have three circuit components - two capacitors of capacitance $C$ each and a resistor of resistance $R$, all of which are connected in parallel. The resistance across any individual capacitor will be $R$, so will it's time constant be $RC$ or something else? I am doubtful that it will remain $RC$ as the second capacitor will also come into the picture.

If I have to find the time constant of such a circuit, do I have to first find the net capacitance ($2C$ in this case) and then multiply it with $R$?
- David White almost 3 years
  
  A picture would help. If all 3 circuit elements are connected in parallel, there will be no resistor in series with either capacitor, and your equation for an RC circuit will not work.
- Yejus almost 3 years
  
  Draw a circuit diagram and apply Krichoff's current and voltage rules for each capacitor. Since all the elements are connected in parallel, the voltages across them will be the same.
- Dylan Rodrigues almost 3 years
  
  i want to do it without the lengthy stuff..in short i want to use thevnin's method
Dylan Rodrigues almost 3 years

there is only one R
trula almost 3 years

sorry I overlooked it: If there is only one resistor then of cause it would be 2RC
Dylan Rodrigues almost 3 years

you combined the two capacitors as one and then calculated the time constant..how should i proceed if i want to find the time constant of each individual capacitor
Dylan Rodrigues almost 3 years

the above comment is precisely my doubt
Bob D almost 3 years

@DylanRodrigues I have updated my answer to respond.