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

  • Option 1)

    2

  • Option 2)

    1

  • Option 3)

    \frac{1}{2}

  • Option 4)

    \frac{1}{4}

 

Answers (1)

As we learnt in 

Self Inductance -

An emf is induced in the coil or the circuit which oppose the change that causes it. Which is also known back  emf.

- wherein

 

 Charge stored at capacitor during discharge is

q=q_{0}. e^{-\frac{t}{RC}}

Energy stored at any time =\frac{q^{2}}{2C}

=\frac{q_0^{2}}{2C}\:e^{-2t/RC}

at t=t_1,\:E=\frac{E_o}{2}=\frac{1}{2}.(\frac{qo^{2}}{2c})\:\frac{1}{2}(\frac{qo^{2}}{2c})=\frac{qo^{2}}{2c}.\:e^{-2t/RC}

taking log. \frac{2t_{1}}{RC}=log_{e}{2}

t_{1}=\frac{RC}{2}log_{e}2-------------------------------(1)

at\ \, t=t_{2},\:q=\frac{qo}{4}=q_{o}.e^{-}\frac{t_{2}}{RC}

t_{2}=2\ln{2}-----------------------------(2)

\therefore \frac{t_{1}}{t_{2}}=\frac{1}{4}

 


Option 1)

2

This is incorrect option

Option 2)

1

This is incorrect option

Option 3)

\frac{1}{2}

This is incorrect option

Option 4)

\frac{1}{4}

This is correct option

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