Question

Find the effective value or rms value (in ampere) of an alternating current in one time period that changes according to the given law: (All quantities are in S.I. unit and symbols have their usual meaning)
$I = 10$, when $0 < t < \dfrac{T}{8}$; $I = 0$, when $\dfrac{T}{8} < t < \dfrac{T}{2}$
$I = - 10$, when $\dfrac{T}{2} < t < \dfrac{5}{8}T$
$I = 0$, when $\dfrac{5}{8}T < t < T$
$I = 10$, when $T < t < \dfrac{9}{8}T$

Answer 1

Hint
The effective value or the rms value of the current in the full time period can be found by integrating the current in the time period $0 < t < T$. So we can find the rms value by integrating the current over the small limits in time period.
In this solution we will be using the following formula,
$\Rightarrow I_{rms}^2 = \dfrac{{\int_0^T {{I^2}dt} }}{{\int_0^T {dt} }}$
where ${I_{rms}}$ is the rms value of current
$I$ is the alternating current
$t$ is the time in the limit $0 < t < T$

Complete step by step answer
To find the effective value or the rms value of current we need to integrate the current in the whole limit of $0 < t < T$. This is given by the formula,
$\Rightarrow I_{rms}^2 = \dfrac{{\int_0^T {{I^2}dt} }}{{\int_0^T {dt} }}$
So in this formula, we can break the limit of the numerator into the smaller parts as given in the question. And for the denominator, the integration will occur under the limit $0 < t < T$.
Therefore, we can write
$\Rightarrow I_{rms}^2 = \dfrac{{\int_0^{{T \mathord{\left/ {\vphantom {T 8}} \right. } 8}} {{I^2}dt} + \int_{{T \mathord{\left/ {\vphantom {T 8}} \right. } 8}}^{{T \mathord{\left/ {\vphantom {T 2}} \right. } 2}} {{I^2}dt} + \int_{{T \mathord{\left/ {\vphantom {T 2}} \right. } 2}}^{{{5T} \mathord{\left/ {\vphantom {{5T} 8}} \right. } 8}} {{I^2}dt} + \int_{{{5T} \mathord{\left/ {\vphantom {{5T} 8}} \right. } 8}}^T {{I^2}dt} }}{{\int_0^T {dt} }}$
In the denominator the value is, $\int_0^T {dt} = \left. t \right|_0^T$
Therefore it gives the value in the denominator as, $\left[ {T - 0} \right] = T$
So we get,
$\Rightarrow I_{rms}^2 = \dfrac{1}{T}\left[ {\int_0^{{T \mathord{\left/ {\vphantom {T 8}} \right. } 8}} {{I^2}dt} + \int_{{T \mathord{\left/ {\vphantom {T 8}} \right. } 8}}^{{T \mathord{\left/ {\vphantom {T 2}} \right. } 2}} {{I^2}dt} + \int_{{T \mathord{\left/ {\vphantom {T 2}} \right. } 2}}^{{{5T} \mathord{\left/ {\vphantom {{5T} 8}} \right. } 8}} {{I^2}dt} + \int_{{{5T} \mathord{\left/ {\vphantom {{5T} 8}} \right. } 8}}^T {{I^2}dt} } \right]$
Now for the numerator, we substitute the values of the current as given in the question. That is,
$\Rightarrow I = 10$, when $0 < t < \dfrac{T}{8}$;
$\Rightarrow I = 0$, when $\dfrac{T}{8} < t < \dfrac{T}{2}$
$\Rightarrow I = - 10$, when $\dfrac{T}{2} < t < \dfrac{5}{8}T$
$\Rightarrow I = 0$, when $\dfrac{5}{8}T < t < T$
So we get,
$\Rightarrow I_{rms}^2 = \dfrac{1}{T}\left[ {\int_0^{{T \mathord{\left/ {\vphantom {T 8}} \right. } 8}} {{{\left( {10} \right)}^2}dt} + \int_{{T \mathord{\left/ {\vphantom {T 8}} \right. } 8}}^{{T \mathord{\left/ {\vphantom {T 2}} \right. } 2}} {{{\left( 0 \right)}^2}dt} + \int_{{T \mathord{\left/ {\vphantom {T 2}} \right. } 2}}^{{{5T} \mathord{\left/ {\vphantom {{5T} 8}} \right. } 8}} {{{\left( { - 10} \right)}^2}dt} + \int_{{{5T} \mathord{\left/ {\vphantom {{5T} 8}} \right. } 8}}^T {{{\left( 0 \right)}^2}dt} } \right]$
The values of the current are constants and come out of the integration, and the time is integrated over the limits.
$I_{rms}^2 = \dfrac{1}{T}\left[ {100\left. t \right|_0^{{T \mathord{\left/ {\vphantom {T 8}} \right. } 8}} + 0 + 100\left. t \right|_{{T \mathord{\left/ {\vphantom {T 2}} \right. } 2}}^{{{5T} \mathord{\left/ {\vphantom {{5T} 8}} \right. } 8}} + 0} \right]$
On substituting the limits we get
$\Rightarrow I_{rms}^2 = \dfrac{1}{T}\left[ {100\left( {\dfrac{T}{8} - 0} \right) + 100\left( {\dfrac{{5T}}{8} - \dfrac{T}{2}} \right)} \right]$
The subtraction in fraction gives us, $\dfrac{{5T}}{8} - \dfrac{T}{2} = \dfrac{{5T - 4T}}{8}$
So it is equal to $\dfrac{{5T - 4T}}{8} = \dfrac{T}{8}$
Substituting we get
$\Rightarrow I_{rms}^2 = \dfrac{1}{T}\left[ {100 \times \dfrac{T}{8} + 100 \times \dfrac{T}{8}} \right]$
We can take the $T$ common and cancel it with the $\dfrac{1}{T}$ outside. So we get
$\Rightarrow I_{rms}^2 = \left[ {\dfrac{{100}}{8} + \dfrac{{100}}{8}} \right]$
This gives us, $I_{rms}^2 = \dfrac{{200}}{8} = 25$
So to get the rms current we take square root on both the sides and get
$\Rightarrow {I_{rms}} = \sqrt {25} = 5A$
Hence the rms value of current is equal to $5A$.

Note
The rms value is the root mean square value of the sinusoidally varying values of current and voltages. The rms value of current and voltages can also be found by dividing the peak to peak value by an amount of $\sqrt 2$.