In the transistor configuration, $I_{1}, I_{2}$ and $I_{3}$ are ammeter readings. Then

I_{1} > I_{2} > I_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{1} < I_{2} < I_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{2} > I_{3} > I_{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

I_{2} > I_{1} > I_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{1} = I_{2} = I_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $I_{2} > I_{3} > I_{1}$
In the given $n - p - n$ transistor, clearly $I_{2} = I_{1} + I_{3}$
Also, in general, the value of $I_{1}$ is much less than $I_{2}$ or $I_{3}$ .
Hence, $I_{2} > I_{3} > I_{1}$

Suggest Corrections

Similar questions

I_{1} = π / 2 \int 0 cos (sin x) d x; I_{2} = π / 2 \int 0 sin (cos x) d x

I_{3} = π / 2 \int 0 cos x d x,

I_{1} = \int_{0}^{\frac{π}{4}} x^{2008} (t a n x)^{2008} d x, I_{2} = \int_{0}^{\frac{π}{4}} x^{2009} (t a n x)^{2009} d x a n d I_{3} = \int_{0}^{\frac{π}{4}} x^{2010} (t a n x)^{2010} d x

I_{1} = \int_{0}^{\frac{π}{2}} c o s (π s i n^{2} x) d x, I_{2} = \int_{0}^{\frac{π}{2}} c o s (2 π s i n^{2} x) d x a n d I_{3} = \int_{0}^{\frac{π}{2}} c o s (π s i n x) d x,

I_{1} = 1 \int 0 e^{- x} {cos}^{2} x d x

I_{2} = 1 \int 0 e^{- x^{2}} {cos}^{2} x d x

I_{3} = 1 \int 0 e^{- x^{3}} d x

Now what should be the correct relation between $I_{1}$ , $I_{2}$ and $I_{3}$ .

Join BYJU'S Learning Program