# Finding time constant of the circuit

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Find the time constant of the circuit after closing the switch .

I tried it by thinking that since the voltage that would be across the 6R would be voltage across the capacitor plates and hence it should
been 6RC but given answer is 3RC

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This can be solved using Thevenin's Theorem :

Any linear electrical network with only voltage and current sources and resistances between its two terminals AB can be replaced by an equivalent voltage source $V_{th}$ in series with an equivalent resistance $R_{th}$.
$V_{th}=V_{oc}$ is the voltage obtained at terminals A-B of the network with terminals A-B open circuited.
$R_{th}$ is the resistance between terminals A and B if all ideal voltage sources are replaced by a short circuit and all ideal current sources are replaced by an open circuit.
If terminals AB are connected to one another, the current $I_{sc}$ flowing between them will be $V_{th}/R_{th}$. This means that $R_{th}$ could alternatively be calculated as $V_{th}$ / $I_{sc}$.

In your circuit we take the terminals AB as being across the capacitor.

To find $R_{th}$ we short-circuit $V$. The $3R$ and $6R$ resistors are then in parallel so they are equivalent to $2R$ because $\frac16+\frac13=\frac12$. This parallel combination is in series with the $R$ resistor so $R_{th}=2R+R=3R$.

The time constant is therefore $3RC$.

It is not required but $V_{th}=V_{oc}$ is the voltage across the $6R$ resistor, which by the voltage divider principle is $\frac{6R}{3R+6R}V=\frac23V$. This is the maximum voltage to which the capacitor will be charged.

References :

answered Apr 5, 2017 by (28,448 points)
selected Apr 6, 2017