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

Q: 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. The the ratio $t _{1} / t _{2}$ will be

d $U\,\, = \,\,\frac{1}{2}C{V^2}\,\,\, \Rightarrow \frac{{{U_0}}}{2}\,\, = \,\,\frac{1}{2}C{V_0}^2{e^{ - 2{t_1}/RC}}\,$ $\, \Rightarrow \frac{1}{2}\,\, = \,\,{e^{ - 2{t_1}/RC}}\,\,\,\,[\because \,\,{U_0}\,\, = \,\,\frac{1}{2}C{V_0}^2]$ $\therefore - \frac{{2{t_1}}}{{RC}}\,\, = \,\,\ln 2\,\,\, \Rightarrow {t_1}\,\, = \,\,\frac{{RC\ln 2}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\,\,(i)\,$ અને $\frac{{{q_0}}}{4}\,\, = \,\,{q_0}{e^{ - {t_2}/RC}}$ $ - \frac{{{t_2}}}{{RC}}\,\, = \,\,2\ln 2\,\,\,\, \Rightarrow {t_2}\,\, = \,\,2RC\ln 2\,\,\,\,\,\,\,\,\,\,\,...........\,\,(ii)$ $[ \frac{{{t_1}}}{{{t_2}}}\,\, = \,\,\frac{1}{4}]$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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. The the ratio $t _{1} / t _{2}$ will be

A$2$
B$1$
C$0.5$
D$0.25$

AIEEE 2010, Diffcult

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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