Capacitor question
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Re: Capacitor questionin this question the capacitor has to be charged up to at least 4.3V before the circuit becomes active this has to be achieved in a time of 60 seconds. we only have learnt the formula for a capacitor being discharged so we can think of the cell as being discharged instead so for a 5V cell that has dissipated 4.3V to a capacitor only 0.7V will be left in the cell so we can use the formula
V=V0e-t/RC
where:
V=0.7
V0=5
t=60
C=2200x10-6
now you can find R you should get a value of 1.39x104
which is approximately 13.9 kilo-ohms -
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Re: Capacitor questionBut if we take initial V to be 5, and final V to be 0.7, then isn't the time delay going to be less than 1 minute? because by the time it gets to 0.7V it is already below the threshold amount(4.3)required for the circuit to be active. Doesn't the capacitance need to have at least 4.3V for the circuit to be active? meaning we must take 4.3 to be final V? Also, how do other members in this forum add special mathematical symbols?(Original post by 99llewellyn99)
in this question the capacitor has to be charged up to at least 4.3V before the circuit becomes active this has to be achieved in a time of 60 seconds. we only have learnt the formula for a capacitor being discharged so we can think of the cell as being discharged instead so for a 5V cell that has dissipated 4.3V to a capacitor only 0.7V will be left in the cell so we can use the formula
V=V0e-t/RC
where:
V=0.7
V0=5
t=60
C=2200x10-6
now you can find R you should get a value of 1.39x104
which is approximately 13.9 kilo-ohmsLast edited by osake; 02-06-2012 at 16:00. -
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Re: Capacitor questionif you look at my post i mentioned that the formula is for discharging only and as the capacitor is charging up it cant be used here however the battery is being discharged and the formula can be used for the battery.(Original post by osake)
But if we take initial V to be 5, and final V to be 0.7, then isn't the time delay going to be less than 1 minute? because by the time it gets to 0.7V it is already below the threshold amount(4.3)required for the circuit to be active. Doesn't the capacitance need to have at least 4.3V for the circuit to be active? meaning we must take 4.3 to be final V? Also, how do other members in this forum add special mathematical symbols?
the final voltage of the battery we assume to be 0.7V when the voltage of the capacitor reaches 4.3V (however this is not possible in reality) therefor i have used 5V as it is the initial voltage of the battery and 0.7V as the final voltage as the battery loses 4.3V to the capacitor in a time of 60 seconds
about the mathematical symbols im not sure -
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Re: Capacitor questionah never knew you could use that formula for batteries. Would the charging formula[Q=Q0(1-e^-t/RC)]work on the capacitor? This formula is from my book but one thing that confuses me about it is, since it is for charging isn't the initial value always 0?(Original post by 99llewellyn99)
if you look at my post i mentioned that the formula is for discharging only and as the capacitor is charging up it cant be used here however the battery is being discharged and the formula can be used for the battery.
the final voltage of the battery we assume to be 0.7V when the voltage of the capacitor reaches 4.3V (however this is not possible in reality) therefor i have used 5V as it is the initial voltage of the battery and 0.7V as the final voltage as the battery loses 4.3V to the capacitor in a time of 60 seconds
about the mathematical symbols im not sure -
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Re: Capacitor questionyes you are nearly right sbout the charging formula and Q0 is initial charge but the formula for charging really should be Q=Qmax(1-e-t/rc)(Original post by osake)
ah never knew you could use that formula for batteries. Would the charging formula[Q=Q0(1-e^-t/RC)]work on the capacitor? This formula is from my book but one thing that confuses me about it is, since it is for charging isn't the initial value always 0?
