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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→Q,B→Q

$${A}\rightarrow{Q},{B}\rightarrow{Q} \\ $$

Commented by rahul 19 last updated on 30/Jun/18

Correct Ans. is A→R , B→Q , C→Q , D→P

$$\mathrm{Correct}\:\mathrm{Ans}.\:\mathrm{is}\: \\ $$$$\mathrm{A}\rightarrow\mathrm{R}\:,\:\mathrm{B}\rightarrow\mathrm{Q}\:,\:\mathrm{C}\rightarrow\mathrm{Q}\:,\:\mathrm{D}\rightarrow\mathrm{P}\: \\ $$

Commented by rahul 19 last updated on 30/Jun/18

(MrW3 , Ajfour) sir pls help us ......

$$\left(\mathrm{MrW3}\:,\:\mathrm{Ajfour}\right)\:\mathrm{sir}\:\mathrm{pls}\:\mathrm{help}\:\mathrm{us}\:...... \\ $$

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

$${q}={q}_{\mathrm{0}} \left(\mathrm{1}−{e}^{−\frac{{t}}{{CR}}} \right)= \\ $$$$\left.{A}\right){energy}=\frac{\mathrm{1}}{\mathrm{2}}\frac{{q}^{\mathrm{2}} }{{c}} \\ $$$$\left.{B}\right){heat}={joules}\:{heat}\:=\frac{{i}^{\mathrm{2}} {Rt}}{{j}} \\ $$$${i}=\frac{{dq}}{{dt}}={q}_{\mathrm{0}} \left(\mathrm{0}−{e}^{\frac{−{t}}{{cR}}} ×\frac{−\mathrm{1}}{{cR}}\right)={q}_{\mathrm{0}} \left(\frac{\mathrm{1}}{{cR}}{e}^{\frac{−{t}}{{cR}}} \right) \\ $$$${heat}=\frac{{q}_{\mathrm{0}} ^{\mathrm{2}} }{{c}^{\mathrm{2}} {R}^{\mathrm{2}} }{e}^{−\mathrm{2}\frac{{t}}{{cR}}} ×{R}×\frac{{t}}{{j}} \\ $$$$\left.{C}\left.\right){energy}_{} =\frac{\mathrm{1}}{\mathrm{2}}{C}×\left({pd}\right)^{\mathrm{2}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{c}\left(\frac{{q}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{2}{c}}{q}_{\mathrm{0}} ^{\mathrm{2}} ×\left(\mathrm{1}−{e}^{\frac{−{t}}{{cR}}} \right)^{\mathrm{2}} \\ $$$$\left.{D}\right){w}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{q}^{\mathrm{2}} }{{c}}=\frac{{q}_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{1}−{e}^{\frac{−{t}}{{cR}}} \right)^{\mathrm{7}} }{\mathrm{2}{c}} \\ $$$$ \\ $$$${energy}=\frac{{q}^{\mathrm{2}} }{\mathrm{2}{c}}\:\:\:\:\:\:\:{q}={q}_{{o}} \left(\mathrm{1}−{e}^{−\frac{{t}}{{cR}}} \right)\approx\left\{\mathrm{1}−\left(\mathrm{1}−\frac{{t}}{{cR}}\right)\right\} \\ $$$${so}\:\:{q}\approx{q}_{\mathrm{0}} \frac{{t}}{{cR}} \\ $$$${energy}=\frac{{q}_{\mathrm{0}} ^{\mathrm{2}} {t}^{\mathrm{2}} }{\mathrm{2}{c}^{\mathrm{3}} {R}^{\mathrm{2}} }\:\:{so}\:{the}\:{graph}\:{alike}\:{y}={x}^{\mathrm{2}} \\ $$$$ \\ $$

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

i am tryiny to corelate with graph

$${i}\:{am}\:{tryiny}\:{to}\:{corelate}\:{with}\:{graph} \\ $$

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