Original post by Tilly-Elizabeth
1.
A capacitor is discharging through a resistor and the time constant is 5.0 s. The time taken for the capacitor to lose half its charge is...?
2.
I have no idea how to work this out.
The time constant is RC.
So you have been given the vale of RC in the equation for the discharge of a capacitor.
Do you know this equation? It includes RC in an exponential.
Original post by Tilly-Elizabeth
1.
A capacitor is discharging through a resistor and the time constant is 5.0 s. The time taken for the capacitor to lose half its charge is...?
2.
I have no idea how to work this out.
It's a little bit tricky but not so difficult when the penny drops.
You know that:
Q=CV and C is constant for any capacitor?
then keeping the equation balanced
0.5Q = 0.5CV
therefore half the charge means half the voltage since C cannot change.
Spoiler
(edited 10 years ago)
Original post by Stonebridge
The time constant is RC.
So you have been given the vale of RC in the equation for the discharge of a capacitor.
Do you know this equation? It includes RC in an exponential.
So you have been given the vale of RC in the equation for the discharge of a capacitor.
Do you know this equation? It includes RC in an exponential.
v=vo e^-t/rc???
Original post by Tilly-Elizabeth
v=vo e^-t/rc???
Yes. Post #3 above has provided you with the answer.
Original post by uberteknik
It's a little bit tricky but not so difficult when the penny drops.
You know that:
Q=CV and C is constant for any capacitor?
then keeping the equation balanced
0.5Q = 0.5CV
therefore half the charge means half the voltage since C cannot change.
You know that:
Q=CV and C is constant for any capacitor?
then keeping the equation balanced
0.5Q = 0.5CV
therefore half the charge means half the voltage since C cannot change.
Spoiler
Original post by Stonebridge
Yes. Post #3 above has provided you with the answer.
Thank you both so much - I understand this now
Original post by Tillybop
1.
A capacitor is discharging through a resistor and the time constant is 5.0 s. The time taken for the capacitor to lose half its charge is...?
2.
I have no idea how to work this out.
Time constant = 5s = RC
Let Q0 =1C therfore we will uae Q = 0.5C as we are looking for time till charge halves
Q = Q0e-t/RC
0.5 = 1e-t/5
ln0.5 = -t/5( ln e)
t = -5( ln0.5)
t = 3.5 (2sf)
Original post by Kishi22
Calculate the time taken for a capacitor to lose half of its charge when the Capacitance is 330μF and resistance is 150kOhm
It's the same as Akeel is worng's post above except you have to change the time constant RC to these values of R and C. Need to put capacitance in farad and resistance in ohm to get time constant in seconds.
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