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What is the effective value (RMS value) of current $i = 10\sin 100\pi t + 10\cos \left( {100\pi t + 30^\circ } \right)$, where $i$ is in ampere?
$\left( A \right)10A$
$\left( B \right)10\sqrt 2 A$
$\left( C \right)10\sqrt 3 A$
$\left( D \right)5\sqrt 2 A$
Answer
370.5k+ views
Hint: We have given the equation of the current at time t. Now from theta equation on further simplification we will get the basic equation of the current in terms of sinusoidal wave form and from that we can get the peak value of the current. Now using the formula we can get the relation between the RMS value and the peak value of the current ad from that we can find our solution.
Complete step by step solution:
As per the problem we have current $i = 10\sin 100\pi t + 10\cos \left( {100\pi t + 30^\circ } \right)$, where i is in ampere.
Now,
$i = 10\sin 100\pi t + 10\cos \left( {100\pi t + 30^\circ } \right)$
Using trigonometric formula,
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Now using this formula in the above equation in the above equation we will get,
$i = 10\sin 100\pi t + 10\left\{ {\cos 100\pi t\cos 30^\circ - \sin 100\pi t\sin 30^\circ } \right\}$
On further simplifying we will get,
$i = 10\sin 100\pi t + 10\left\{ {\dfrac{{\sqrt 3 }}{2}\cos 100\pi t - \dfrac{1}{2}\sin 100\pi t} \right\}$
$ \Rightarrow i = 10\sin 100\pi t - \dfrac{{10}}{2}\sin 100\pi t + 10\left\{ {\dfrac{{\sqrt 3 }}{2}\cos 100\pi t} \right\} = \dfrac{{10}}{2}\sin 100\pi t + 10\dfrac{{\sqrt 3 }}{2}\cos 100\pi t$
Taking $10$ as common we will get,
$ \Rightarrow i = 10\left( {\dfrac{1}{2}\sin 100\pi t + \dfrac{{\sqrt 3 }}{2}\cos 100\pi t} \right)$
Now converting $\dfrac{1}{2}$ and $\dfrac{{\sqrt 3 }}{2}$ in trigonometric form we will get,
$ \Rightarrow i = 10\left( {\cos 30^\circ \sin 100\pi t + \sin 30^\circ \cos 100\pi t} \right)$
Now using trigonometric formula,
$\cos A\sin B + \sin A\cos B = \sin \left( {A + B} \right)$
Now using this formula we will get,
$ \Rightarrow i = 10\sin \left( {100\pi t + 30^\circ } \right)$
The basic equation of current is given as,
$i = {I_0}\sin \left( {\omega t + \phi } \right)$
Now on comparing we will get,
${I_0} = 10$
Now RMS value formula we will get,
${I_{RMS}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$
Now putting the known value we will get,
${I_{RMS}} = \dfrac{{10}}{{\sqrt 2 }} = \dfrac{{10\sqrt 2 }}{2} = 5\sqrt 2 A$
So, the correct answer is “Option D”.
Note: Remember that the RMS value is defined as the effective value of a varying voltage or current. We can calculate the RMS current by dividing the RMS voltage by the impedance of the circuit. Note that the RMS value is the equivalent steady DC value which gives the same effect.
Complete step by step solution:
As per the problem we have current $i = 10\sin 100\pi t + 10\cos \left( {100\pi t + 30^\circ } \right)$, where i is in ampere.
Now,
$i = 10\sin 100\pi t + 10\cos \left( {100\pi t + 30^\circ } \right)$
Using trigonometric formula,
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Now using this formula in the above equation in the above equation we will get,
$i = 10\sin 100\pi t + 10\left\{ {\cos 100\pi t\cos 30^\circ - \sin 100\pi t\sin 30^\circ } \right\}$
On further simplifying we will get,
$i = 10\sin 100\pi t + 10\left\{ {\dfrac{{\sqrt 3 }}{2}\cos 100\pi t - \dfrac{1}{2}\sin 100\pi t} \right\}$
$ \Rightarrow i = 10\sin 100\pi t - \dfrac{{10}}{2}\sin 100\pi t + 10\left\{ {\dfrac{{\sqrt 3 }}{2}\cos 100\pi t} \right\} = \dfrac{{10}}{2}\sin 100\pi t + 10\dfrac{{\sqrt 3 }}{2}\cos 100\pi t$
Taking $10$ as common we will get,
$ \Rightarrow i = 10\left( {\dfrac{1}{2}\sin 100\pi t + \dfrac{{\sqrt 3 }}{2}\cos 100\pi t} \right)$
Now converting $\dfrac{1}{2}$ and $\dfrac{{\sqrt 3 }}{2}$ in trigonometric form we will get,
$ \Rightarrow i = 10\left( {\cos 30^\circ \sin 100\pi t + \sin 30^\circ \cos 100\pi t} \right)$
Now using trigonometric formula,
$\cos A\sin B + \sin A\cos B = \sin \left( {A + B} \right)$
Now using this formula we will get,
$ \Rightarrow i = 10\sin \left( {100\pi t + 30^\circ } \right)$
The basic equation of current is given as,
$i = {I_0}\sin \left( {\omega t + \phi } \right)$
Now on comparing we will get,
${I_0} = 10$
Now RMS value formula we will get,
${I_{RMS}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$
Now putting the known value we will get,
${I_{RMS}} = \dfrac{{10}}{{\sqrt 2 }} = \dfrac{{10\sqrt 2 }}{2} = 5\sqrt 2 A$
So, the correct answer is “Option D”.
Note: Remember that the RMS value is defined as the effective value of a varying voltage or current. We can calculate the RMS current by dividing the RMS voltage by the impedance of the circuit. Note that the RMS value is the equivalent steady DC value which gives the same effect.
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