Question
Everything is on the picture. Questions are:
3. Has i) ii) and iii) parts to it.
Use graph for reference. Ignore the vertical lines drawn on the graph at t=19 and t=37 (I was obviously lost).
4. Has a) and b) part to it.
A capacitor, which has a capacitance of 20,000 muFD, has an initial potential difference of 2.4V across it. The experiment is allowed to run for about 80 seconds. The estimated timing error is 2 seconds. The following plot shows the variation ot voltage with time as the capacitor is being discharged: On the graph, show clearly how the first half-life can be measured. Show also how the second half -life can be measured. Do the hall- lives measured in (a) and (b) agree? Show why. A capacitor of 0.05 F was used in series with a resistance to study the charging of its voltage. If the half-life for charge was found to be 50 s, calculate the resistance used in the circuit. If the voltage applied to the capacitor in the above charging circuit at, t =0 is 10 V,find the voltage across C at t = 20 s.
Explanation / Answer
First half life is at V=1.2 volt because inital voltage is 2.4 volt so T1=21 sec
Second half life is at V=0.6 vold (half of 1.2) so T2 = 43 sec
the half lifes agree because error in time is 2 sec
Half life=RC/ln2
R=50*ln2*0.05=1.73 Ohm
V=V0*e^(-t/RC)
V=10*e^(-20/50) [RC=time constant]
=6.7 Volt