Transcript Motion Along a Straight Line at Constant Acceleration

Book Reference : Pages 98-101
1.
To qualitatively understand how a capacitor
discharges through a resistor
2.
To derive the equation which defines this
rate of discharge
3.
To be able to solve capacitor discharge
problems
When a charged capacitor is allowed to discharge
through a fixed resistor it does so gradually until
it reaches 0
Charge
Discharge
Switch
C
V0
Fixed
resistor R
We can compare this discharge with water leaving a
tank through a pipe at the bottom, initially the flow
rate is high because of the pressure. At the level falls so
does the pressure reducing the flow rate
Look at the shape of the graphs qualitatively.
They both show curves which starts at the Y axis
and decay asymptotically towards the X axis
The first graph
shows charge from
Q=CV
[Virtual Physics Lab]
The second graph
shows current from
I = V/R
Consider charge, if we start at a
charge of Q0, then after a certain
time t the charge will decay to
say only 0.9Q0 (arbitrary choice)
Experimentally, we can show
that after a further time t the
charge has decayed to
0.9 x 0.9 Q0 after 2 t
and 0.9 x 0.9 x 0.9Q0 after 3 t
and 0.9n Q0 after time nt
The decay is exponential
Exponential decays....
If a quantity decreases at a rate which is proportional to
the quantity (left) then the decay is exponential
Explaining the decrease
Consider one small step in the decay process where Q
decays to Q - Q in a time t
The current at this time is given by
I = V/R
I = Q/CR
from Q=CV, V = Q/C so
From I = Q/t if t is very small then the drop in charge Q can be rewritten as -It and I is therefore -Q/t
Substituting into our earlier equation for I = Q/CR
Q/t = -Q/CR
For infinitely short time intervals as t tends to 0 (t0)
Q/t represents the rate of change of charge & is
written as the first differential dQ/dt hence
dQ/dt = -Q/CR
Solution by integration :
Q = Q0 e–t/RC
Where Q0 is the initial charge & e is
the exponential function
From Q=CV voltage is proportional to charge, similarly
from Ohm’s law Current is proportional to voltage.
All three quantities decay in exactly the same way :
Charge
Q = Q0 e–t/RC
Voltage
V = V0 e–t/RC
Current
I = I0 e–t/RC
The quantity “RC” is called the time constant & is the
time for the initial charge/voltage/current to fall to 0.37
of the initial value (0.37 = e-1)
The units for RC are the second
A 2200F capacitor is charged to a pd of 9V and then
allowed to discharge through a 100k resistor. Calculate
The initial charge on the capacitor
The time constant for the circuit
The pd after a time equal to the time constant
The pd after 300s
The initial charge on the capacitor
Using Q=CV, the initial charge Q0 is 2200F x 9V
= 0.02 C
The time constant for the circuit
Time constant = RC = 100,000 x 2200F = 220s
The pd after a time equal to the time constant
By definition t = RC when V = V0e-1 = 0.37 x 9V = 3.3V
The pd after 300s
Using V = V0 e–t/RC
-t/RC = 300/220 = 1.36 (no units)
V = 9 e-1.36
V = 2.3V
A 50F capacitor is charged by connecting it to a 6V
battery & then discharging it through a 100k resistor.
Calculate :
The initial charge stored [300C]
The time constant for the circuit [5.0s]
Estimate how long the capacitor would take to discharge
to about 2V [5s]
Estimate the size of the resistor required in place of the
100k if 99% of the discharge is to be complete in about
5s [20k]
A 68F capacitor is charged by connecting it to a 9V
battery & then discharging it through a 20k resistor.
Calculate :
The initial charge stored [0.61C]
The initial discharge current [0.45mA]
The pd and the discharge current 5s after the start
of the discharge [0.23V, 11A]