Question

90% பயனுறுதிறன் கொண்ட 200V / 120V இறக்கு மின்மாற்றி ஒன்று 40 Ω மின்தடை கொண்ட மின்தூண்டல் அடுப்புடன் இணைக்கப்பட்டுள்ளது. மின்மாற்றியின் முதன்மைச்சுருளில் பாயும் மின்னோட்டத்தைக் காண்க.

Answer 1

Given,

$\displaystyle\small\longrightarrow \tan^{-1}(2)+\tan^{-1}(3)+x=\pi$

Since $\displaystyle\small\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\dfrac{a+b}{1-ab}\right),$

$\displaystyle\small\longrightarrow \tan^{-1}\left(\dfrac{2+3}{1-2\cdot3}\right)+x=\pi$

$\displaystyle\small\longrightarrow \tan^{-1}\left(-1\right)+x=\pi$

We know $\displaystyle\small\tan^{-1}(-1)=-\dfrac{\pi}{4}$ but here,

$\displaystyle\small\longrightarrow\tan^{-1}(2)\in\left(0,\ \dfrac{\pi}{2}\right)$

$\displaystyle\small\longrightarrow\tan^{-1}(3)\in\left(0,\ \dfrac{\pi}{2}\right)$

So,

$\displaystyle\small\longrightarrow \tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}(-1)\in(0,\ \pi)$

$\displaystyle\small\Longrightarrow\tan^{-1}(-1)\in\left(\dfrac{\pi}{2},\ \pi\right)$

$\displaystyle\small\Longrightarrow\tan^{-1}(-1)=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$

Therefore,

$\displaystyle\small\longrightarrow\dfrac{3\pi}{4}+x=\pi$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{x=\dfrac{\pi}{4}}} }$

Answer 2

Given,

$\displaystyle\small\longrightarrow \tan^{-1}(2)+\tan^{-1}(3)+x=\pi$

Since $\displaystyle\small\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\dfrac{a+b}{1-ab}\right),$

$\displaystyle\small\longrightarrow \tan^{-1}\left(\dfrac{2+3}{1-2\cdot3}\right)+x=\pi$

$\displaystyle\small\longrightarrow \tan^{-1}\left(-1\right)+x=\pi$

We know $\displaystyle\small\tan^{-1}(-1)=-\dfrac{\pi}{4}$ but here,

$\displaystyle\small\longrightarrow\tan^{-1}(2)\in\left(0,\ \dfrac{\pi}{2}\right)$

$\displaystyle\small\longrightarrow\tan^{-1}(3)\in\left(0,\ \dfrac{\pi}{2}\right)$

So,

$\displaystyle\small\longrightarrow \tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}(-1)\in(0,\ \pi)$

$\displaystyle\small\Longrightarrow\tan^{-1}(-1)\in\left(\dfrac{\pi}{2},\ \pi\right)$

$\displaystyle\small\Longrightarrow\tan^{-1}(-1)=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$

Therefore,

$\displaystyle\small\longrightarrow\dfrac{3\pi}{4}+x=\pi$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{x=\dfrac{\pi}{4}}} }$

Answer 3

Given,

$\displaystyle\small\longrightarrow \tan^{-1}(2)+\tan^{-1}(3)+x=\pi$

Since $\displaystyle\small\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\dfrac{a+b}{1-ab}\right),$

$\displaystyle\small\longrightarrow \tan^{-1}\left(\dfrac{2+3}{1-2\cdot3}\right)+x=\pi$

$\displaystyle\small\longrightarrow \tan^{-1}\left(-1\right)+x=\pi$

We know $\displaystyle\small\tan^{-1}(-1)=-\dfrac{\pi}{4}$ but here,

$\displaystyle\small\longrightarrow\tan^{-1}(2)\in\left(0,\ \dfrac{\pi}{2}\right)$

$\displaystyle\small\longrightarrow\tan^{-1}(3)\in\left(0,\ \dfrac{\pi}{2}\right)$

So,

$\displaystyle\small\longrightarrow \tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}(-1)\in(0,\ \pi)$

$\displaystyle\small\Longrightarrow\tan^{-1}(-1)\in\left(\dfrac{\pi}{2},\ \pi\right)$

$\displaystyle\small\Longrightarrow\tan^{-1}(-1)=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$

Therefore,

$\displaystyle\small\longrightarrow\dfrac{3\pi}{4}+x=\pi$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{x=\dfrac{\pi}{4}}} }$