Question

A 12.0 μF capacitor is charged to a potential of 50.0 V and discharged
through a 175 Ω resistor. How long does it take the
capacitor to lose half its charge? A 12.0 μF capacitor is charged to a potential of 50.0 V and discharged
through a 175 Ω resistor. How long does it take the
capacitor to lose half its charge? @Physics

Answer 1

I'd be interested to see the answer to this.

Answer 2

the answer is 1.5*10^-3 seconds, but I don't know how to calculate it, it involves the equation for a charging capacitor. \[\xi/R*e^(-t/RC)\]

Answer 3

The function to calculated voltage is:
\[\Large {V_c}\left( t \right) = {V_i}{e^{ - \frac{t}{{RC}}}}\]
You can work backwards from this to find the time. Step by step:
\[\Large \begin{array}{l}
{V_c}\left( t \right) = {V_i}{e^{ - \frac{t}{{RC}}}}\\
\ln \left( {{V_c}} \right) = \ln \left( {{V_i}{e^{ - \frac{t}{{RC}}}}} \right) = \ln \left( {{V_i}} \right) + \ln \left( {{e^{ - \frac{t}{{RC}}}}} \right)\\
\ln ({V_c}) = \ln ({V_i}) - \frac{t}{{RC}}\\
\ln ({V_c}) - \ln ({V_i}) = - \frac{t}{{RC}}\\
\ln \left( {\frac{{{V_c}}}{{{V_i}}}} \right) = - \frac{t}{{RC}}\\
t = - RC \cdot \ln \left( {\frac{{{V_c}}}{{{V_i}}}} \right)\\
t = - 175 \cdot \left( {12 \times {{10}^{ - 6}}} \right) \cdot \ln \left( {\frac{{25}}{{50}}} \right)\\
= 1.4556 \times {10^{ - 3}}{\rm{s}}
\end{array}\]
A handy formula to keep around would be:
\[\Large t = - RC \cdot \ln \left( p \right)\]Where p is the decimal percentage of voltage change (i.e. Vc/Vi).