Hello. I\'m in the blank spot with this lab and so I don\'tknow how to answer th
ID: 1723183 • Letter: H
Question
Hello. I'm in the blank spot with this lab and so I don'tknow how to answer these questions. Can someone please help me answer this and explain thoroughlyso that I can understand it too?? Please, Thank you in advance and I will also rate you alifesaver! Thanks! (:- In the circuit, R=5.00k,C=35.0 µF and the voltage across the capacitor is 5.00 voltswhen the switch is thrown to B. What is the time constant for thecircuit?
- How long does it take for thevoltage across the capacitor to decrease to one volt?
- Show that you have someunderstanding of logarithms by proving that we don’t actuallyneed to take the separate logarithms of V1 and V2, and that we canjust calculate RC as follows: RC=(t2-t1)/ln(V1/V2)
- In the circuit, R=5.00k,C=35.0 µF and the voltage across the capacitor is 5.00 voltswhen the switch is thrown to B. What is the time constant for thecircuit?
- How long does it take for thevoltage across the capacitor to decrease to one volt?
- Show that you have someunderstanding of logarithms by proving that we don’t actuallyneed to take the separate logarithms of V1 and V2, and that we canjust calculate RC as follows: RC=(t2-t1)/ln(V1/V2)
Explanation / Answer
Well the time constant is RC, which is resistance*capacitance. The first thing you need is the equation for the rate of dischargeof a charged capacitor which is: Q=Ce^-t/RC Q/C=Voe^-t/RC V=Voe^-t/RC thus solve for t using the fact that V is one volt less than theintial. V/Vo=e^-t/RC -ln(V/Vo)RC=t solve for t -ln(5/4)(5x10^3)(35x10^-6)=t t is also time from t=0 so it may be represented t2-t1 so you cansolve for RC RC=-(t2-t1)/ln(V2/V1) -ln(a/b)=ln(b/a) RC=(t2-t1)/ln(V1/V2) Further Clarification Ok, the formula sometimes is easier to simply accept but here itgoes. The first step is understanding that the voltage is comingfrom the potential difference within the capacitor which will beequivalent to the voltage drop through the circuit which as this isan RC circuit only consists of a resistor outside the capacitorwhich gets us V(r)=-V(c) thus from Ohm's Law and the properties ofcapacitors IR=-Q/C The rate at which charge leaves a capacitor is the definition ofcurrent however we can also write this as the derivative dQ/dt sowe put that back in our formula so dQ/dtR=-Q/C dQ/Q=-dt/RC this is simply using arithmetic and differential equations at thispoint, but I did it in a way where I subsituted dQ/dt for I andthen simply divided each side by R and multiplied each side by dtand Q. Now integrate both sides from the point when t=0 to t whichfor the left side of the equation is from the intial charge Qo toQ. so on the left it is ln(Q)-ln(Qo) which is ln(Q/Qo) while theright side is simply going to be -(t-to)/RC. So now we take bothsides to the power of e to get rid of the natural logarithm so wenow have Q/Qo=e^(-t/RC) Q=Qoe^(-t/RC) So we have a formula for the charge across the capacitor but weneed the Voltage so we use the formula V=Q/C so we have V=Voe^(-t/RC) Now to find t simplyf irst solve for that formula above for RC sowe divide each side by Vo and take the natural logarithm ofboth sides and you get ln(V/Vo)=-t/RC RCln(V/Vo)=-t Ok now plug in the values for V=4 since its 1 less volts and Vo at5 volts For the second part solve for RC -t/ln(V/Vo)=RC remember ln(a/b) = -ln(b/a) THIS is your "logarithmknowledge" t/ln(Vo/V)=RC Now if you remember correctly t is actually form theintegral from to to t so t becomes (t2-t1) and there you go.
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