Question

Find the value of \[\dfrac{t}{\tau }\] for which the current in an LR circuit builds up to
(a) 90%
(b) 99% and,
(c) 99.9% of the steady-state value.

Answer 1

Hint: RL circuit may be a series combination of the resistor and inductor. This circuit is connected with a voltage source like a battery and a switch. We use this circuit within the chokes of luminescent tubes and also use this for supplying DC power to Radio Frequency amplifiers.

Complete step-by-step solution:
We know that \[i = {i_0}\left( {1 - {e^{ - \dfrac{t}{\tau }}}} \right)\]
(a) \[i = 90\% i\]
\[\dfrac{{90}}{{100}}i = {i_0}\left( {1 - {e^{ - \dfrac{t}{\tau }}}} \right)\]
\[0.9 = 1 - {e^{ - \dfrac{t}{\tau }}}\]
\[{e^{ - \dfrac{t}{\tau }}} = 0.1\]
Taking log on both sides
\[\ln {e^{ - \dfrac{t}{\tau }}} = \ln 0.1\]
\[\dfrac{t}{\tau } = 2.3\]
(b) \[i = 99\% i\]
\[\dfrac{{99}}{{100}}i = {i_0}\left( {1 - {e^{ - \dfrac{t}{\tau }}}} \right)\]
\[0.99 = 1 - {e^{ - \dfrac{t}{\tau }}}\]
\[{e^{ - \dfrac{t}{\tau }}} = 0.01\]
Taking log on both sides
\[\ln {e^{ - \dfrac{t}{\tau }}} = \ln 0.01\]
\[\dfrac{t}{\tau } = 4.6\]
(c) \[i = 99.9\% i\]
\[\dfrac{{99.9}}{{100}}i = {i_0}\left( {1 - {e^{ - \dfrac{t}{\tau }}}} \right)\]
\[0.999 = 1 - {e^{ - \dfrac{t}{\tau }}}\]
\[{e^{ - \dfrac{t}{\tau }}} = 0.001\]
Taking log on both sides
\[\ln {e^{ - \dfrac{t}{\tau }}} = \ln 0.001\]
\[\dfrac{t}{\tau } = 6.9\]
Hence, the value of \[\dfrac{t}{\tau }\] for which the current in an LR circuit builds up to \[90\% ,99\% \] and \[99.9\% \] of the steady-state value is \[2.3,4.9 \text{ and } 6.9\] respectively.

Note: A resistor–inductor circuit may be an electrical circuit composed of resistors and inductors driven by a voltage source. A first-order RL circuit consists of one resistor and one inductor. It is the simplest type of RL circuit. A primary order RL circuit is one in every of the simplest analog infinite impulse response electronic filters. It consists of a resistor and an inductor, either asynchronous driven by a voltage source or in parallel driven by a current source.