Question

$\text{The Question : }$

The Initial Phase Angle For Current $\sf I = 10 \sin \omega t+8 \cos \omega t,$ is

$\text{Option : }$

1. $\tan^{-1} \bigg(\frac{4}{5}\bigg)$

2. $\tan^{-1} \bigg( \frac{5}{3}\bigg)$

3. $\sin^{-1} \bigg(\frac{4}{5}\bigg)$

4. $\sf 90°$

Answer Is (1) $\tan^{-1} \bigg(\frac{4}{5}\bigg)$

Need Explanation ✔️ Please

Please Don't spam ❌

​

Answer 1

Answer:

1

Explanation:

First let simply a little

$i = 10 \sin( \omega t) + 8 \cos( \omega t) \\ \\ = \sqrt{10 {}^{2} + {8}^{2} } \bigg\{ \frac{10}{ \sqrt{ {10}^{2} + {8}^{2} } } \sin( \omega t) + \frac{8}{ \sqrt{ {10}^{2} + {8}^{2} } }{ \cos \omega t} \bigg \}$

Let

$\sin( \alpha ) = \frac{8}{ \sqrt{164} } \: \: then \: \: \cos( \alpha ) = \frac{10}{ \sqrt{164} }$

So we get

$i = { \sqrt{164} } \{\cos( \alpha ) \sin( \omega t) + \sin( \alpha ) \cos(\omega t) \} \\ \\ i = { \sqrt{164} } \sin(\omega t + \alpha )$

Clearly initial phase angle is alpha . Now

$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{8}{10} = \frac{4}{5} \\ \\ \alpha = \tan {}^{ - 1} ( \frac{4}{5} )$

Answer 2

Explanation:

Revise the concept of SHM

This type of equation can be solved to give the desired values associated with it.

Suppose the equation is,

$\rm y = asin\omega t + bcos\omega t$

Now, let $\rm a = Acos\phi ,\ b = Asin\phi$

Now, the equation becomes as:

$\rm :\longmapsto y = Acos\phi sin\omega t + A sin\phi cos\omega t$

$\rm :\longmapsto y = A(sin\omega t cos\phi + cos\omega t sin\phi )$

$\rm :\longmapsto y=Asin(\omega t + \phi )$

Hence, we can deduce that A = amplitude and $\phi$ = phase difference

Now dividing the equations of a, b:

$\rm :\longmapsto \dfrac{b}{a} = \dfrac{Asin\phi}{Acos\phi}$

$\rm :\longmapsto \dfrac{b}{a} = tan\phi$

$\rm :\longmapsto \phi = tan^{-1}\dfrac{b}{a}$

Hence we got the value of phase difference

Now considering the given equation in the question, we get

$\rm :\longmapsto a = 10$

$\rm :\longmapsto b = 8$

Hence phase angle is given by,

$\rm :\longmapsto \phi = tan^{-1}\dfrac{b}{a}$

$\rm :\longmapsto \phi = tan^{-1}\dfrac{\cancel{8}^4}{\cancel{10}_5}$

$\boxed{\boxed{\rm :\longmapsto \phi = tan^{-1}\dfrac{4}{5}}}$

$\color{#FF7968}{\rule{36pt}{7pt}}\color{#4FB3F6}{\rule{36pt}{7pt}}\color{#FBBE2E}{\rule{36pt}{7pt}}\color{#60D399}{\rule{36pt}{7pt}}\color{#6D83F3}{\rule{36pt}{7pt}}$