Question

at what time after the discharge begins is the energy stored in the capacitor reduced to half its initial value?

Answer 1

The time at which the energy stored in the capacitor will reduce to half of its initial value is 0.00693 s.

The time at which the energy stored in the capacitor will reduce to half of its initial value is calculated as follows;

$Q = Q_o e^{-\frac{t}{RC} }$

• the time constant, RC = 10 ms = 0.01 s
• Q₀ is the initial charged stored
• Q is the final charge of the capacitor

$\frac{Q_0}{2} = Q_0 e^{-\frac{t}{RC} }\\\\\frac{Q_0}{2Q_0} = e^{-\frac{t}{RC} }\\\\\frac{1}{2} = e^{-\frac{t}{RC} }\\\\ln(\frac{1}{2} ) = -\frac{t}{RC} \\\\ln(\frac{1}{2} ) = - \frac{t}{0.01} \\\\-0.693 = - \frac{t}{0.01}\\\\t = 0.00693 \ s$

Thus, the time at which the energy stored in the capacitor will reduce to half of its initial value is 0.00693 s.

"Your question is not complete, it seems to be missing the following information;"

"The capacitor in an RC circuit is discharged with a time constant of 10.0ms. At what time after the discharge begins is the charge on the capacitor reduced to half its initial value."

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