Question 322: Electrostats & Capacitance

Question

Let $$C$$ be the capacitance of a capacitor discharging through a resistor $$R.$$ Suppose $${t_1}$$ is the time taken for the energy stored in the capacitor to reduce  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 $${t_1}/{t_2}$$ will be

Accepted Answer

Initial energy of capacitor, $${E_1} = {{q_1^2} \over {2C}}$$
Final energy of capacitor, $${E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} ight)^2}$$
$$	herefore$$ $${t_1}=$$  time for the charge to reduce to $${1 \over {\sqrt 2 }}$$ of its initial value
and $${t_2} = $$ time for the charge to reduce to $${1 \over 4}$$ of its initial value
We have, $${q_2} = {q_1}{e^{ - t/CR}}$$
$$ \Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} ight) =  - {t \over {CR}}$$
$$	herefore$$$$\ln \left( {{1 \over {\sqrt 2 }}} ight) = {{ - {t_1}} \over {CR}}...\left( 1 ight)$$
and $$\ln \left( {{1 \over 4}} ight) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 ight)$$
By $$(1)$$ and $$(2),$$ $${{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} ight)} \over {\ln \left( {{1 \over 4}} ight)}}$$
$$ = {1 \over 2}{{\ln \left( {{1 \over 2}} ight)} \over {2\ln \left( {{1 \over 2}} ight)}} = {1 \over 4}$$

Answer

$$1$$

Answer

$${1 \over 2}$$

Answer

$${1 \over 4}$$

Answer

$$2$$

Electrostats & Capacitance

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