Question

The current in an L_R circuit builds up to 3/4th of its steady state value in 4 seconds. Find the time constant of this circuit.

Answer 1

$\huge \mathfrak{solution}$

Given:

• The current in an LR circuit builds up to 3/4th of its steady current .

• t= 4 seconds

To find :

• Time constant =?

Answer:

As we know that :

$\star \boxed{\bold{ i = i_o(1 - {e}^{ \frac{ - t}{ \tau} } ) }}$

Here:

$i = \frac{3}{4} i_o$

$\implies \: \frac{3}{4} i_o = i_o(1 - {e}^{ \frac{ - 4}{ \tau} } ) \\ \\ \implies \frac{3}{4} \cancel{i_o} = \cancel{i_o}(1 - {e}^{ \frac{ - 4}{ \tau} } ) \\ \\ \implies \: \frac{3}{4} = 1 - {e}^{ \frac{ - 4}{ \tau} } \\ \\ \implies \: {e}^{ \frac{ - 4}{ \tau} } = 1 - \frac{3}{4} = \frac{1}{4} \\ \\ \implies \: {e}^{ \frac{ - 4}{ \tau} } = \frac{1}{4}$

Take Log both sides:

$\implies \: log( {e}^{ \frac{ - 4}{ \tau} } ) = log( \frac{1}{4} ) \\ \\ \implies \: \frac{ - 4}{ \tau} log(e) = log(1) - log(4) \: \: \: \boxed{ log(e) = 1} \: \boxed{ log(1) = 0} \\ \\ \implies \frac{ - 4}{ \tau} = - log(4) \\ \\ \implies \: \frac{ - 4}{ \tau} = - log( {2}^{2} ) = - 2 log(2) \\ \\ \implies \tau = \frac{2}{ log(2) }$

So, The value of time constant is 2/Ln2

Answer 2

Explanation:

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