Thevenin's theorem on Capacitor.

Question

The time constant of the circuit shown is:

I found $R_{th}=\dfrac{3R}{5}$ by following method :

We short-circuit the batteries to calculate Thevenin's equivalent resistance across $C$ hence $2R\parallel 2R=R$ which is parallel with $\dfrac{R}{2}+R=\dfrac{3R}{2}\implies R_{th}=R\parallel \dfrac{3R}{2}=\dfrac{3R}{5}$
So time constant $\tau=\dfrac{3RC}{5}$

But $\tau=\dfrac{RC}{2}$, given. Where I went wrong?

Answer 1

The mistake you have made is to replace the EMF labelled $2E$ with an open circuit. If you replace it with a short-circuit then the vertical resistance $R$ and the internal resistance $R$ are both shorted out - there is no resistance between F and G.

It is a good idea to re-draw the circuit after shorting the EMFs :

The EMF labelled $E, R$ cannot be shorted because it appears to have an internal resistance of $R$. There is no indication that the EMF labelled $2E$ has any internal resistance. When this is shorted there is no resistance between the points F and G. Points F, G, H are then at the same potential, so the 3 resistors $R$ (horizontal), $2R$ (diagonal) and $2R$ (vertical) are in parallel.

The resistance between A and B is therefore $\frac12 R$.

Comments

If potential difference across a resistor is zero then we don't consider that, right?

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If the potential difference across a resistor is zero then there is a short-circuit. It has zero resistance. It is the same as an ideal wire.