A total charge Q flows across a resistor R during a time interval T in such a way that the current vs. time graph for 0 to T is like the loop of a sin curve in the range 0 to $π$ . The total heat generated in the resistor is

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$Q^{2} π^{2} R / 4 T$

$2 Q^{2} π^{2} R / T$

$2 Q^{2} π^{2} R / 4 T$

$Q^{2} π^{2} R / 8 T$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$I = I_{0} \sin ω t$

$Q = \int Idt = I_{0} \int_{0}^{T / 2} \sin ω tdt = I_{0} {[\frac{- \cos ω t}{ω}]}_{0}^{T / 2}$

$= \frac{I_{0}}{(2 π / T)} [- \cos ω t]_{0}^{T / 2} = \frac{I_{0} T}{π}$

$\Rightarrow I_{0} = \frac{Q π}{T} \dots . (1)$

Total heat $H = \int I^{2} Rdt = R \int {(I_{0} \sin ω t)}^{2} dt$

$= I_{0}^{2} R \int_{0}^{T / 2} (\frac{1 - \cos 2 ω t}{2}) dt$

$= \frac{I_{0}^{2} R}{2} {[t - \frac{\sin 2 ω t}{2 ω}]}_{0}^{T / 2}$

$= \frac{I_{0}^{2} R}{2} [T / 2 - 0 - 0 + 0]$

$= \frac{I_{0}^{2} RT}{4} = \frac{Q^{2} π^{2}}{T^{2}} \times \frac{RT}{4} \Rightarrow H = \frac{Q^{2} π^{2} R}{4 T} .$

Watch 3-min video & get full concept clarity