Capacitance alevel physics help
Capacitance Resistance
100μF 200kΩ
Part A
Time constant
a) What is the time constant?
Value 20 s
What is the halving time?
How do I work this out, I thought it would be 10s
100μF 200kΩ
Part A
Time constant
a) What is the time constant?
Value 20 s
What is the halving time?
How do I work this out, I thought it would be 10s
Reply 1
Original post
by LisaPereraCapacitance Resistance
100μF 200kΩ
Part A
Time constant
a) What is the time constant?
Value 20 s
What is the halving time?
How do I work this out, I thought it would be 10s
100μF 200kΩ
Part A
Time constant
a) What is the time constant?
Value 20 s
What is the halving time?
How do I work this out, I thought it would be 10s
The key concept:
The time constant τ (tau) and half-life are two different ways of characterising the same exponential process, but they are not simply half of each other.
Why it is not simply τ/2:
During the discharge of a capacitor, the voltage (or charge) follows this law:
V(t) = V° · e^(-t/τ)
The half-life t1/2 is the time required for the voltage to drop to half its initial value; therefore:
V°/2 = V°· e^(-t1/2/τ)
The calculation process:
Divide both sides by V°:
1/2 = e^(-t1/2/τ)
Take the natural logarithm of both sides:
ln(1/2) = -t1/2/τ
Remember that ln(1/2) = -ln(2):
-ln(2) = -t1/2/τ
Solve fort1/2:
t1/2 = τ · ln(2)
Therefore:
The half-life is the time constant multiplied by ln(2) ≈ 0.693.
In your case: t1/2 = 20 s × 0.693 ≈ 13.9 s
It is not 10 s because exponential decay is not linear – at the beginning, the capacitor discharges faster, then slower and slower!
Ciao,
Sandro
(edited 2 months ago)
Reply 2
12
Original post
by NitrotolueneLet me explain how to find the half-life!
The key concept:
The time constant τ (tau) and half-life are two different ways of characterising the same exponential process, but they are not simply half of each other.
Why it is not simply τ/2:
During the discharge of a capacitor, the voltage (or charge) follows this law:
V(t) = V° · e^(-t/τ)
The half-life t1/2 is the time required for the voltage to drop to half its initial value; therefore:
V°/2 = V°· e^(-t1/2/τ)
The calculation process:
Divide both sides by V°:
1/2 = e^(-t1/2/τ)
Take the natural logarithm of both sides:
ln(1/2) = -t1/2/τ
Remember that ln(1/2) = -ln(2):
-ln(2) = -t1/2/τ
Solve fort1/2:
t1/2 = τ · ln(2)
Therefore:
The half-life is the time constant multiplied by ln(2) ≈ 0.693.
In your case: t1/2 = 20 s × 0.693 ≈ 13.9 s
It is not 10 s because exponential decay is not linear – at the beginning, the capacitor discharges faster, then slower and slower!
Ciao,
Sandro
The key concept:
The time constant τ (tau) and half-life are two different ways of characterising the same exponential process, but they are not simply half of each other.
Why it is not simply τ/2:
During the discharge of a capacitor, the voltage (or charge) follows this law:
V(t) = V° · e^(-t/τ)
The half-life t1/2 is the time required for the voltage to drop to half its initial value; therefore:
V°/2 = V°· e^(-t1/2/τ)
The calculation process:
Divide both sides by V°:
1/2 = e^(-t1/2/τ)
Take the natural logarithm of both sides:
ln(1/2) = -t1/2/τ
Remember that ln(1/2) = -ln(2):
-ln(2) = -t1/2/τ
Solve fort1/2:
t1/2 = τ · ln(2)
Therefore:
The half-life is the time constant multiplied by ln(2) ≈ 0.693.
In your case: t1/2 = 20 s × 0.693 ≈ 13.9 s
It is not 10 s because exponential decay is not linear – at the beginning, the capacitor discharges faster, then slower and slower!
Ciao,
Sandro
Thank you so much
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