R અવરોધમાંથી ડિસ્ચાર્જિંગ થતાં કેપેસિટરનું કે | vedclass

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Question

C is the capacitance, $R$ is the resistor, $t$ is time taken.

The function of time is given by

$U =\frac{1}{2} \frac{ q ^2}{ c }$ Where $u$ is t

Vedclass · Accepted Answer

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Vedclass · Answer

C is the capacitance, $R$ is the resistor, $t$ is time taken.

The function of time is given by

$U =\frac{1}{2} \frac{ q ^2}{ c }$ Where $u$ is the initial velocity, $q$ is the charge.

The charges are the function of time

$q=q_0 e^{\frac{-t}{r}}$

Where $T$ is the relaxation time,

So,

$T=R c$

$U =\frac{1}{2} \frac{ q _0^2 e ^{\frac{-2 t }{ T }}}{ c }$

$U=U_0 e^{\frac{-2 t}{T}}$

at $t = t _1, U =\frac{ U _0}{2}$

$\frac{ U _0}{2}= U _0 e ^{\frac{-2 t }{ T }}$

$\frac{1}{2}=e^{\frac{-2 t}{T}}$

Taking log both sides

$t _1=\frac{ T }{2} \log (2)$

Now,

$q=q_0 e^{\frac{-t}{T}}$

At $t = t { }^2$

$q=\frac{q_0}{4}$

$\frac{q_0}{4}=q_0 e^{\frac{-t_2}{T}}$

$\frac{1}{4}=e^{\frac{-t_2}{T}}$

Again, taking log both sides,

$t _2=2 T \log (2)$

Now,

$\frac{ t _1}{ t _2}=\frac{\frac{ T }{2} \log (2)}{2 T \log (2)}$

$\frac{t_1}{t_2}=\frac{1}{4}$

Thus, the ratio will be $\frac{ t _1}{ t _2}=\frac{1}{4}$

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