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Question Number 38675 by rahul 19 last updated on 28/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18

$A \to Q, B \to Q$
Commented by rahul 19 last updated on 30/Jun/18

$Correct Ans . is$ $A \to R, B \to Q, C \to Q, D \to P$
Commented by rahul 19 last updated on 30/Jun/18

$(MrW 3, Ajfour) sir pls help us . . . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18
	$q=q_0 (1−e^(−(t/(CR))) )= A)energy=(1/2)(q^2 /c) B)heat=joules heat =((i^2 Rt)/j) i=(dq/dt)=q_0 (0−e^((−t)/(cR)) ×((−1)/(cR)))=q_0 ((1/(cR))e^((−t)/(cR)) ) heat=(q_0 ^2 /(c^2 R^2 ))e^(−2(t/(cR))) ×R×(t/j) C)energy_ =(1/2)C×(pd)^2 } =(1/2)×c((q^2 /c^2 ))=(1/(2c))q_0 ^2 ×(1−e^((−t)/(cR)) )^2 D)w=(1/2)×(q^2 /c)=((q_0 ^2 (1−e^((−t)/(cR)) )^7 )/(2c)) energy=(q^2 /(2c)) q=q_o (1−e^(−(t/(cR))) )≈{1−(1−(t/(cR)))} so q≈q_0 (t/(cR)) energy=((q_0 ^2 t^2 )/(2c^3 R^2 )) so the graph alike y=x^2$
	$q = q_{0} (1 - e^{- \frac{t}{C R}}) =$ $A) e n e r g y = \frac{1}{2} \frac{q^{2}}{c}$ $B) h e a t = j o u l e s h e a t = \frac{i^{2} R t}{j}$ $i = \frac{d q}{d t} = q_{0} (0 - e^{\frac{- t}{c R}} \times \frac{- 1}{c R}) = q_{0} (\frac{1}{c R} e^{\frac{- t}{c R}})$ $h e a t = \frac{q_{0}^{2}}{c^{2} R^{2}} e^{- 2 \frac{t}{c R}} \times R \times \frac{t}{j}$ $C) {e n e r g y}_{} = \frac{1}{2} C \times {(p d)}^{2}}$ $= \frac{1}{2} \times c (\frac{q^{2}}{c^{2}}) = \frac{1}{2 c} q_{0}^{2} \times {(1 - e^{\frac{- t}{c R}})}^{2}$ $D) w = \frac{1}{2} \times \frac{q^{2}}{c} = \frac{q_{0}^{2} {(1 - e^{\frac{- t}{c R}})}^{7}}{2 c}$ $e n e r g y = \frac{q^{2}}{2 c} q = q_{o} (1 - e^{- \frac{t}{c R}}) \approx {1 - (1 - \frac{t}{c R})}$ $s o q \approx q_{0} \frac{t}{c R}$ $e n e r g y = \frac{q_{0}^{2} t^{2}}{2 c^{3} R^{2}} s o t h e g r a p h a l i k e y = x^{2}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18

	$i a m t r y i n y t o c o r e l a t e w i t h g r a p h$
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