Question 55: AC fundamentals

Question

The figure given below shows an
LCR
series circuit with two switches S
1
and S
2
. When switch S
1
is closed keeping S
2
open, the phase difference (φ) between the current and source voltage is 30° and phase difference is 60° when S
2
is closed keeping S
1
open. The value of (3L
1
− L
2
) is ______ H.
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1mnrr8zzu/d21d648a-f575-42e1-95ae-3fb8d8ee270d/1b9d09a0-343a-11f1-87a1-6b3ff65a2683/file-6y3zli1mnrr8zzv.png?format=png)

Accepted Answer

The root mean square (r.m.s) value of a time-varying current $i(t)$ over a period $T$ is defined by the formula :
$ \mathrm{i}_{\mathrm{rms}}=\sqrt{\frac{1}{\mathrm{~T}} \int\limits_0^{\mathrm{T}} \mathrm{i}^2 \mathrm{dt}} $
We are given $\mathrm{i}=\mathrm{i}_{\mathrm{o}}\left(\frac{\mathrm{t}}{\mathrm{T}}ight)$.
So, using the rms formula,
$ \mathrm{i}_{\mathrm{rms}}^2=\frac{1}{\mathrm{~T}} \int\limits_0^{\mathrm{T}}\left(\mathrm{i}_{\mathrm{o}}^2 \frac{\mathrm{t}^2}{\mathrm{~T}^2}ight) \mathrm{dt} $
Since $i_o$ and $T$ are constants, we can move them outside the integral :
$\mathrm{i}_{\mathrm{rms}}^2=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3} \int_0^{\mathrm{T}} \mathrm{t}^2 \mathrm{dt}=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3}\left[\frac{\mathrm{t}^3}{3}ight]_0^{\mathrm{T}}=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3} 	imes \frac{\mathrm{T}^3}{3}$
$\Rightarrow \mathrm{i}_{\mathrm{rms}}=\sqrt{\frac{\mathrm{i}_{\mathrm{o}}^2}{3}}$
$\Rightarrow $ $ \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{3}} $
Hence, the correct option is (B).

Answer

$\dfrac{9}{2}$

Answer

$\dfrac{2}{9}$

Answer

$\dfrac{1}{3}$

Answer

3

AC fundamentals

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