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

11 i_{0}^{2} R t_{0}

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

13 i_{0}^{2} R t_{0}

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

17 i_{0}^{2} R t_{0}

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

15 i_{0}^{2} R t_{0}

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

Open in App

Solution

The correct option is B $13 i_{0}^{2} R t_{0}$
For first part
$\begin{matrix} i {(\frac{3 i_{o}}{t_{o}})}^{t} \int d H = \int i^{2} R d t \Rightarrow H = R \int_{0}^{t_{o}} \frac{9 {i_{o}}^{2}}{{t_{o}}^{2}} . t^{2} d t = \frac{9 R {i_{o}}^{2}}{{t_{o}}^{2}} \frac{t^{3}}{3} = 3 {i_{o}}^{2} R t_{o} ∴ T o t a l h e a t = 3 {i_{o}}^{2} R t_{o} + {(3 i_{o})}^{2} R t_{o} + {i_{o}}^{2} R t_{0} = 13 {i_{o}}^{2} R t_{o} H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

t_{0}

R

T

q

R

t

Q = a t - b t^{2}

R

Join BYJU'S Learning Program