# How to find $\tau$ in this circuit?

1)

2)

I know, that $$τ = \frac{L}{R}$$ but what is $R$ in this formula? It seems to be the total resistance, but how to find it in 1) and 2)?

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## 2 Answers

Have derived it myself:

$${\tau}=\dfrac{\underbrace {\large { I}}_{\text{through inductor at steady state}}}{\underbrace {{dI_{\small }}/{dt}}_{\text{initial}}}$$

Also $$V_{\text{inductor}}=L\dfrac{dI}{dt}$$

where $I$ is current through inductor.

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This answer would be much improved if you show the derivation. –  David Z Jun 3 '13 at 19:12

but what is R in this formula?

Since you're interested in what happens after the switch opens in both cases, redraw the circuit after the switch opens.

In (1), there is just the one resistor R to the left of the switch so that's the resistance in the time constant.

In (2), there is just the two series connected 1k resistors so R = 2k is the resistance to use.

If the formula for $\tau$ given by user007 seems mysterious, here's a derivation.

Assume, for concreteness, an RL circuit with initial current $i_0$. The current is given by:

$i(t) = i_0 \cdot e^{-t / \tau}, t \ge 0$

where

$\tau = L/R$

Now, take the time derivative of both sides to get:

$\frac{di(t)}{dt} = i_0 \cdot e^{-t / \tau}(-1/\tau)$

Then, evaluate this at $t = 0$:

$\frac{di(0)}{dt} = i_0 \cdot e^{-0 / \tau}(-1/\tau) = -i_0/\tau$

Thus:

$\tau = \dfrac{i_0}{\frac{di(0)}{dt}}$

(the negative sign is required in this case since the current is decaying, i.e., the time rate of change of current is negative)

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