Question

In a LR circuit discharging current is given by $I = {I_0}{e^{ - \dfrac{t}{\tau }}}$ where $\tau$ is the time constant of the circuit. Find the RMS current from the period $t = 0$ to $t = \pi$.

Answer 1

Hint- A LR circuit is basically a circuit having inductors and resistors. Inductors having Inductance L and the resistors of resistance R are connected in series. The time constant of the particular LR circuit is defined as the ratio of inductance and resistance connected in that circuit. And the time constant describes the growth or decay of current in the LR circuit. And the RMS value denotes the amount of current that dissipates the same power in the resistor.

Formula used:
$I_{rms}^2 = \dfrac{1}{\tau }\int\limits_0^\tau {{I^2}} dt$
Where,
${I_{rms}}$= RMS current that dissipated
$\tau$=Time constant

Complete step by step answer:
(i)In the question, the value of current that dissipates in a time T is given as, $I = {I_0}{e^{ - \dfrac{t}{\tau }}}$ applying this current $I$value in the formula,
$I_{rms}^2 = \dfrac{1}{\tau }\int\limits_0^\tau {{I^2}} dt$
$\therefore I_{rms}^2 = \dfrac{1}{\tau }\int_0^\tau {I_o^2} {e^{\dfrac{{ - 2t}}{\tau }}}dt$
(ii) Now integrating the above equation with the limit 0 to$\tau$,
$I_{rms}^2 = I_o^2\left( { - \dfrac{\tau }{2}e_\tau ^{ - t}} \right)$
$\rightarrow \dfrac{{I_o^2}}{\tau }\left( {\dfrac{\tau }{2}\left( {1 - {e^{ - 1}}} \right)} \right)$
$\rightarrow \dfrac{{I_o^2}}{2}\left( {1 - {e^{ - 2}}} \right)$
$\therefore {I_{rms}} = \dfrac{{{I_o}}}{{\sqrt 2 }}\sqrt {\left( {1 - {e^{ - 2}}} \right)}$
Hence the RMS current in between the time period 0 to $\tau$is,$\therefore {I_{rms}} = \dfrac{{{I_o}}}{{\sqrt 2 }}\sqrt {\left( {1 - {e^{ - 2}}} \right)}$
(iii) The RMS current means the root mean square value of a current. It is the amount of current that dissipates the power in a resistor. The RMS value of overall time of a periodic function is equivalent to the one period of that function.
(iv)The RMS current is equal to the ratio of peak current ${I_o}$divided by$\sqrt 2$.
$\rightarrow {I_{rms}} = \dfrac{{{I_o}}}{{\sqrt 2 }}$

Note: RMS value of current and voltage are very important. Because the AC voltage current and voltages keep on changing. If we want to build an equivalent AC circuit for the particular DC circuit means we can use the RMS value of current and voltage. It is the same for the overall period and one period of any function. The RMS values are always lesser than the peak value of the particular quantity.