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# Answer to Question #84095 in Electric Circuits for Oliver Jones

Question #84095
An example Capacitor is 24v and 1500&mu;F and has a time constant of 0.15 Seconds. I need to determine (using a discharge equation) the total charge, in coulombs discharged in the first 5 time constants and then predict how long it would take to discharge to 0.03 coulombs.
1
2019-01-09T09:49:05-0500

The discharge equation:

@$V(t)=V_0e^{-t/T},@$

where @$T=RC@$ is a time constant. The total charge in the first 5 time constants thus is:

@$Q=C\cdot V(5T)=C\cdot V_0e^{-5}= @$

@$=1.5\cdot10^{-3}\cdot24\cdot e^{-5}=2.43\cdot10^{-4} \textrm{ C}.@$

Now predict how long it would take to discharge our capacitor to 0.03 coulombs (when the voltage of course will be @$V=Q/C=0.03/(1.5\cdot10^{-3})=20 \textrm{ V}@$). Express @$t@$ from the first formula:

@$t=-RC\cdot {\textrm{ln}\frac{V}{V_0}}=-0.15\cdot {\textrm{ln}\frac{20}{24}}=0.027\textrm{ s}@$

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