You visited us 2 times! Enjoying our articles? Unlock Full Access!

Transistor

A time varyin...

Question

$A$ time varying current $i$ is passed through a resistance $R$ as shown in figure. The total heat generated in the resistance is

13 (i_{0})^{2} T t_{0}

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

11 (i_{0})^{2} T t_{0}

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

15 (i_{0})^{2} T t_{0}

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

17 (i_{0})^{2} T t_{0}

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

Open in App

Solution

The correct option is A $13 (i_{0})^{2} T t_{0}$
From $t = 0 \to t_{0}$
From the graph
$i = \frac{3 i_{0}}{i_{0}} t$ (from straight line equation)

We know that, $H = i^{2} R t^{2}$
Total heat produced
$= \int_{0}^{t_{0}} (\frac{3 i_{0}}{t_{0}})^{2}$ $R d t + (3 i_{0})^{2} R (2 t_{0} - t_{0}) + (i_{0})^{2} R (3 t_{0} - 2 t_{0})$
$= 3 (i_{0})^{2} R t_{0} + 9 (i_{0})^{2} R t_{0} + (i_{0})^{2}$ $R t_{0} = 13 (i_{0})^{2} R t_{0}$
Hence, option $(b)$ is correct.

Suggest Corrections

Similar questions

t_{0}

R

t

Q = a t - b t^{2}

R

E = 10 V

R

t_{1}

t_{2}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program