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 ${t_2}$ is the time taken for the charge to reduce to one fourth its initial value. Then the ratio $\dfrac{{{t_1}}}{{{t_2}}}$ will be

Answer 1

A resistor circuit containing both inductor and capacitor, the current will take some time to attain its maximum peak value. If the battery is removed from the circuit which is having a capacitor or an inductor, then the current takes some time to decay to zero value. Use the above statement to determine the given ratio.

Formula used:
$q = {q_0}{e^{\frac{{ - t}}{c}}}$
q is the charge, $t$ is the time.

Complete step by step answer:
If the DC source is one in a resistance containing circuit, then the current will attain its maximum steady value in zero-time interval. A resistor circuit containing both inductor and capacitor, the current will take some time to attain its maximum peak value. If the battery is removed from the circuit which is having a capacitor or an inductor, then the current takes some time to decay to zero value.
Resonance occurs in a circuit that is when the inductor, capacitor and resistor are connected in series when the supply frequency causes the voltage across the inductor and capacitor to be equal. Q factor will be affected if there is resistive loss. Q factor is a unit less dimensionless quantity.
The initial velocity is given, $U = \dfrac{1}{2}\dfrac{{{q^2}}}{c}........\left( 1 \right)$ here u is the initial velocity, q is the charge.
The charge is the function of time
$q = {q_0}{e^{\frac{{ - t}}{c}}}$
Here $T = RC$
Also, $T$ is the relaxation time.
Now we have to putting the values in $\left( 1 \right)$, we get
$U = \dfrac{1}{2}\dfrac{{q_0^2{e^{\frac{{ - 2t}}{T}}}}}{C}....\left( 2 \right)$
Take $t = {t_1},U = \dfrac{{{U_0}}}{2}$
$\Rightarrow \dfrac{{{U_0}}}{2} = {U_0}{e^{\frac{{ - 2{t_1}}}{T}}}$
Taking log on both sides and we get
$\Rightarrow {t_1} = \dfrac{T}{2}\log \left( 2 \right)$
Now we take $t = {t_2}$ and $U = \dfrac{{{q_0}}}{4}$in $\left( 2 \right)$and we get
$\Rightarrow \dfrac{{{q_0}}}{4} = {q_0}{e^{\frac{{ - {t_2}}}{T}}}$
Taking log on both sides we get
$\Rightarrow {t_2} = 2T\log \left( 2 \right)$
Now we have to find out the ratio of ${t_1}$ and ${t_2}$we get
$\Rightarrow \dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{\dfrac{T}{2}\log \left( 2 \right)}}{{2T\log \left( 2 \right)}}$
On cancel the term and we get
$\Rightarrow \dfrac{{{t_1}}}{{{t_2}}} = \dfrac{1}{4}$

Note: If the battery is removed from the circuit which is having a capacitor or an inductor, then the current takes some time to decay to zero value. Q factor is the energy stored per unit cycle to energy dissipated per cycle. Quality factor controls the damping of oscillations. It is a dimensionless quantity.