Question

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t₁ is the time taken for the energy stored in the capacitor to reduce 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 of a capacitor discharging through a resistor R.

• t₁ is the time taken for the energy stored in the capacitor to reduce to half its initial value

• t₂ is the time taken for the charge to reduce to one-fourth its initial value.

⇝ To Find :-

• Value of t₁ / t₂

⇝ Solution :-

❒ We Know equation of discharging of a charge is :

$\bf q = q_{ \circ} {e}^{-t/RC} \: - - - (1)$

★ According To Question :

$\pink {\text{At time t = t₂ }}$

$\text q = \dfrac{\text q_{ \circ}}{4}$

✏ Using (1) :

$:\longmapsto\dfrac {\cancel{\text q_{ \circ}}}{4} = \cancel\text q_{ \circ} {\text e}^{ - { \text t_2/ \text{RC}}}$

$:\longmapsto\sf\dfrac{1}{4} = {e}^{-t_2/RC}$

★ Taking Log Both Side :

$:\longmapsto \sf \log \bigg( \dfrac{1}{4} \bigg) = \log \big( {e}^{-t_2/RC} \big)$

$:\longmapsto \sf \log {(4)}^{ - 1} = \log( {e)}^{-t_2/RC}$

$:\longmapsto \sf \cancel- 1( \log4) =\cancel - \dfrac{t_2}{RC}. \log \: e$

$:\longmapsto \sf \log4 = \dfrac{t_2}{RC}$

$:\longmapsto \sf t_2 = RC .\log4$

$:\longmapsto \sf t_2 = RC . \log{2}^{2}$

$:\longmapsto \boxed{ \pink{ \boxed{\bf t_2 = 2\times RC \times log2}}}$

❒As Energy Stored in a capacitor is :

$\text E = \frac{1}{2} \dfrac{ {\text q_{ \circ}}^{2} }{ \text{C}} \: \: - - - (2)$

★ Let when charge is q₁ , Energy becomes E/2

So,

$\sf \dfrac{E}{2} = \frac{1}{2} \frac{q_1 {}^{2} }{C}$

✏ Using (2) :

$:\longmapsto \sf \cancel\dfrac{1}{2} \frac{ {q_{ \circ}}^{2} }{2 \cancel C } =\cancel\dfrac{1}{2} \frac{ {q_1}^{2} }{C}$

$:\longmapsto \bf q_1 = \dfrac{q_{ \circ}}{\sqrt{2}}\:\: ----(3)$

✏ Using (1) :

$:\longmapsto \sf q_1 = q_{ \circ} { \: e}^{ - t_1/RC}$

✏ Using (3) :

$:\longmapsto \sf \dfrac{ \cancel {q_{ \circ}}}{ \sqrt{2} } = \cancel{q_{ \circ}} \: {e}^{ - t_1/RC}$

$:\longmapsto \sf \dfrac{1}{ \sqrt{2} } = {e}^{ - t_1/RC}$

★ Taking Log Both Side :

$:\longmapsto \sf\log \bigg( \dfrac{1}{ \sqrt{2} } \bigg) = \log( {e)}^{ - t_1/RC}$

$:\longmapsto \sf\log(2) {}^{ - 1/2} = \log( {e)}^{ - t_1/RC}$

$:\longmapsto \sf \cancel - \dfrac{1}{2} \log(2) = \cancel - \dfrac{t_1}{RC} \log \: e$

$:\longmapsto \sf\dfrac{1}{2} \log2 = \dfrac{t_1}{RC}$

$:\longmapsto \boxed{ \pink{ \boxed{\bf t_2 = \frac{1}{2} \times RC \times log2}}}$

Hence,

$\sf \dfrac{t_1}{t_2} = \frac{\frac{1}{2} \times \cancel{RC} \times \cancel{\log2}} { 2 \times \cancel{RC} \times \cancel{\log2}}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf  \frac{t_1}{t_2} = \frac{1}{4} } }}}$

Answer 2

$\color{red}{}REFER \: \: THE \: \: ATTACHMENT \: \: FOR \: \: ANSWER$