Question

Let C be the capacitance of a capacitor discharging through a resistor R.Suppose t is the time taken for the energy stored in the capacitor to be reduced to half its initial value and t2 is the time taken for the charge to reduce to one fourth its initial value.Then the ratio t1/t2 will be

Answer 1

Given,

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,

$T=Rc$

$U=\frac{1}{2}\frac{q_0^2{e}^{\frac{-2t}{T}}}{c}$

$U=U_0{e}^{\frac{-2t}{T}}$

at $t=t_1,U=\frac{U_0}{2}$

$\frac{U_0}{2}=U_0{e}^{\frac{-2t}{T}}$

$\frac{1}{2}=e^{\frac{-2t}{T}}$

Taking log both sides

$t_1=\frac{T}{2}log\left(2\right)$

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,

$t_2=2Tlog\left(2\right)$

Now,

$\frac{t_1}{t_2}=\frac{\frac{T}{2}log\left(2\right)}{2Tlog\left(2\right)}$

$\frac{t_1}{t_2}=\frac{1}{4}$

Thus, the ratio will be $\frac{t_1}{t_2}=\frac{1}{4}$