8. Let\'s say you measure t1/2 to be 0.0163 seconds and the resistance in the ci

Question

8. Let's say you measure t1/2 to be 0.0163 seconds and the resistance in the circuit is 94 Ohms. Then what is the capacitance in micro Farads?

Capacitance =__________ micro Farads. (Answer with three significant figures or N/A if there is no answer.)

(t1/2 means that the voltage (or charge) of the system will increase to half more of what is left in a time equal to t1/2 seconds. Therefore if a system is already at half charge (t1/2 seconds after starting) then after t1/2 more seconds the system will be charged to 50% plus half of 50%. That is 25% more, or 75% of the entire charge.Let's say that four t1/2's have gone by. That means that the charge (or voltage) is at (50% + 1/2*50% + 1/2*1/2*50% + 1/2*1/2*1/2*50%) = 93.75% of maximum charge. Yikes! Now look in your manual for a more simple mathematical derivation of this concept.)

9. Given t1/2 to be 0.0163 seconds, how long should it take to reach 75% of maximum charge? time = _________seconds. (Answer with three significant figures or N/A if there is no answer.)

I have worked these several times over and can't seem to get the proper answer! Thank you for your help.

Explanation / Answer

8) Q = Qo*[1 - e^(-t/RC)]

0.5 = [1 - e^(-0.0163/94*C)] ===> e^(-0.0163/94*C) = 0.5

  (-0.0163/94*C) = -0.693

C = 2.5*10^-4 F

9)  "Therefore if a system is already at half charge (t1/2 seconds after starting) then after t1/2 more seconds the system will be charged to 50% plus half of 50%. That is 25% more, or 75% of the entire charge."
Surely that tells us that the time for 75% charge is 2*time for 50% charge = 2*0.0163 = 0.0326 seconds.

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