# Time constant of this circuit

Can someone tell me on why the time constant of the below circuit is not R2*C?

simulate this circuit – Schematic created using CircuitLab

My understanding is that, the Capacitor C1 is getting charged only through R2 and I feel that R1 isn't playing any role. I tried to find answer in this site and got this link. But I am not sure on the logic why we should calculate the time constant with the parallel combination of the resistors.

How to go about finding the Time constant of this circuit intuitively?

I tried to reduce the resistor network, but not sure on how to proceed.

• electronics.stackexchange.com/questions/377467/…
– G36
Commented Oct 20, 2021 at 18:06
• If R1 was not in circuit, the voltage on C would go to 10V. But the voltage divider means the voltage on C is limited to voltage across R1. Commented Oct 20, 2021 at 20:48
• R1 and R2 both have some effect on the time constant. Therefore, you cannot ignore R1. The rule is short out DC sources when calculating time constants. Once you short out V1, you will see that they really are in parallel. Commented Oct 21, 2021 at 5:22
• You may think of R1 as a discharge circuit for C. So it does play a role. Commented Oct 21, 2021 at 19:15

## Intuitively

What follows is completely lacking in rigour, but you asked for intuitive, and here it is. I'm going to make these observations:

• R1 forms a potential divider with R2. If the capacitor were not there, the potential at the junction of R1 and R2 would be: $$10V \times \frac{R_1}{R_1+R_2} = 3\frac{1}{3}V = 3.33V$$ With C1 in place, this voltage will be C1's final, charged voltage.

• C1's rate of charge is proportional to the current through it.

• C1 is initially discharged, with 0V across it. Therefore R1 also has 0V across it, and consequently no current passes through R1, initially.

• All current through R2 must therefore initally pass through C1, and this current will be the same in both cases (the full 10V across R2).

• Initial rate of charge of C1 will be the same in both cases.

• When R1 is present, the final "target" voltage is one third of the case when R1 is absent. Since the initial rate of change of voltage across C1 is the same in both cases, the capacitor will reach that target in one third of the time, with R1 in place.

That value of "one third" is key. It comes from the term $$\\frac{R_1}{R_1+R_2}\$$. I cannot think of an intuitive way of relating this to the combined parallel resistance R1 ∥ R2, except to point out:

$$R_2 \times \frac{R_1}{R_1+R_2} = \frac{R_1 R_2}{R_1 + R_2} = R_1 \parallel R_2$$

I confess I can't see how this helps, but at least you can see how the time constant is related to this parallel combination.

## By Thevenin's Theorem

We can replace the entire network consisting of V1, R1 and R2 with its Thevenin equivalent circuit:

simulate this circuit – Schematic created using CircuitLab

$$\R_{TH}\$$ and $$\V_{TH}\$$ are calculated as follows:

\begin{aligned} V_{TH} &= V_1 \frac{R_1}{R_1 + R_2} \\ \\ R_{TH} &= R_1 \parallel R_2 = \frac{R_1R_2}{R_1 + R_2} \end{aligned}

These two circuits are functionally identical. If you enclosed them in a box, with only nodes A and B exposed, there's no way from the outside of distinguishing one from the other, because they behave identically. Any conditions you impose at nodes A and B of either box will yield exactly the same results.

So, if you connected a capacitor C1 between A and B, it would charge in the same way in both cases. Yet, in the second Thevenin equivalent circuit, clearly the time constant will be:

\begin{aligned} \tau &= R_{TH} \cdot C1 \\ \\ &= (R1 \parallel R2) \cdot C1 \end{aligned}

Basically you need to determine the thevenin equivalent resistance which feeds the capacitor.

Intuitively, the capacitor only becomes significant for AC type signals and under AC that voltage source can be viewed as a short-circuit. With the voltage source a short-circuit R1 and R2 are in parallel

Intuitively, there is a 6k resistor that tries to charge the capacitor to 10V, but at the same time there is a 3k resistor that tries to charge the capacitor to 0V.

It would just be more obvious if for example they were both 6k resistors and both would charge to separate 10V supplies.

Calculate thevenin equivalent of DC source and the resistor divider as it is the output of resistor divider that charges the capacitor.

Time constants in linear circuits, in the case of a single reactive component C or L, are by definition calculated in this way:

1. Turn off all of the voltage and current sources

2. Calculate the Thevenin's resistance seen by C

3. Tau = Req * C

A "no brainer" solution method would be to use some mathematical software like Maple to help you solve node equations and inverse-LaPlace transform. Call the voltage at the top of $$\C_1 \$$ for $$\V_C \$$. The node equation is

$$\frac{V_C-V_{\text{in}}}{R_2}+\frac{V_C}{R_1}+V_CsC_1=0$$ where we have LaPlace transformed the capacitor. Solving for $$\V_C \$$ gives $$V_C = \frac{R_1}{sC_1R_2R_1+R_1+R_2}\cdot V_{\text{in}}$$ Assuming that $$\V_{\text{in}}\$$ is a step $$\\frac{10}{s} \$$ inverse-Laplace transforming gives $$V_C = 10\text{V}\cdot\frac{R_1}{R_1+R_2}\cdot \bigg(1-e^{-\frac{R_1+R_2}{C_1R_1R_2}t} \bigg)$$ Comparing the expression to the "standard" RC-charging formula $$\V_C=V_s \cdot K \cdot \bigg(1-e^{\frac{t}{\tau}} \bigg) \$$ which suggests that $$\\tau = \frac{C_1R_1R_2}{R_1+R_2} \$$.

1. Set the voltage source to zero and assume that the capacitor is loaded.
2. Find the resistive network for discharging the capacitor - in your case: R1||R2.
3. T=(R1||R2)*C1.

Short out the capacitor. The current through the cap is obviously 1.667mA. That's what happens when the power is first applied (assuming an initial condition of 0V across the cap).

Now open the cap. The open circuit voltage is 3.333V. That's your final voltage.

So the equivalent charging resistance must be Req = 3.333V/0.001667A charging to a final voltage of 3.333V.

That resistance times the capacitance will give you the time constant.

My understanding is that, the Capacitor C1 is getting charged only through R2 and I feel that R1 isn't playing any role

Let's say that is true: so when the capacitor $$\C_1\$$ gets charged, there will be a voltage across it and since $$\R_1\$$ is in parallel there will be voltage across it too; and consequently a part of the current will be increasingly flowing through $$\R_1\$$. The only case when $$\R_1\$$ would not matter is if it were $$\\infty\$$ or an open circuit or in other words not there at all. Only then will your assertion be valid.

In the settled state with a DC source, there will be no current through the capacitor. The voltage across $$\C_1\$$ will therefore be 10. (3/(3+6)) = 10/3 = 3.33 V.

The parallel impedance of $$\C_1\$$ and $$\R_1\$$ $$Z_1=\frac{R_1}{\left(C_1 s\right) \left(\frac{1}{C_1 s}+R_1\right)}$$

which on simplifying is: $$\frac{R_1}{C_1 R_1 s+1}$$

For ac analysis, the voltage across $$\C_1\$$ would be given by the following ratio,

$$\frac{Z_1}{R_2+Z_1}$$ which upon simplifying becomes:

$$\frac{R_1}{C_1 R_2 R_1 s+R_1+R_2}$$

and putting in the standard pole-zero form:

$$\frac{R_1}{\left(R_1+R_2\right) \left(\frac{C_1 \left(R_1 R_2\right) s}{R_1+R_2}+1\right)}$$

suggesting that the time constant is: $$\\tau =\frac{C_1 \left(R_1 R_2\right)}{R_1+R_2}\$$

• At high frequencies, the $$\s\$$ term will dominate the denominator and the response across $$\C_1\$$ will go to zero.

• At DC, the $$\s\$$ term will be zero, and you will get the simple voltage division ratio.

Here is the intuitive way to think about it that I have recently developed for myself...

Consider two circuits, your example with and without R1 resistor. In both cases the current at t=0 when the capacitor acts like a short the current is dictated by R2 resistor. This is the max current that can charge the capacitor.

So the circuit without R1 has to charge the cap to 10V using this max current (10V/6K ohm). Now adding R1 into the circuit creates a voltage divider at the cap changing the final voltage level it will reach, is still has to charge the cap using this same max current, but now it only charges to 3.33V. Given the same current it will charge to the lower voltage (circuit with R1) in a shorter time than it takes to charge to a higher voltage (circuit without R1).