how a bleeder resistor (R = 260 k) is used to discharge a capacitor (C = 85.0 µF) after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock.
Figure 20.47
(a) What is the time constant?
(b) How long will it take to reduce the voltage on the capacitor to 0.350% of its full value once discharge begins?
(c) If the capacitor is charged to a voltage V0 through a 125 resistance, calculate the time it takes to rise to 0.865V0 (this is about two time constants).
I got part a and b... but i don't know how to get part c! a is 22.1s, part b is 125 s....
To calculate the time it takes for the capacitor to rise to 0.865V0, we can use the formula for the charging of a capacitor through a resistor:
t = 2 * time constant = 2RC
Given that the resistance is 125 Ω and the capacitance is 85.0 µF, we can substitute these values into the formula:
t = 2 * (125 Ω) * (85.0 µF)
First, we need to convert the microfarads to farads by dividing by 1,000,000:
t = 2 * (125 Ω) * (85.0 * 10^(-6) F)
Now, we can calculate the time:
t = 2 * (125 Ω) * (85.0 * 10^(-6) F)
t = 0.02125 seconds
Therefore, it takes approximately 0.02125 seconds (or 21.25 milliseconds) for the capacitor to rise to 0.865V0, which is about two time constants.
To answer part (c) of the question, we need to calculate the time it takes for the capacitor to rise to 0.865V0, where V0 is the initial voltage on the capacitor.
To understand how to approach this, it is important to know the equation that describes the voltage on a capacitor during charging or discharging:
V(t) = V0 * (1 - e^(-t/RC))
Where:
- V(t) is the voltage on the capacitor at time t
- V0 is the initial voltage on the capacitor
- t is the time elapsed since charging/discharging began
- R is the resistance in the circuit
- C is the capacitance of the capacitor
For part (c), we want to find the time it takes for V(t) to reach 0.865V0, which we can represent as:
0.865V0 = V0 * (1 - e^(-t/RC))
Now, we can rearrange this equation to find t:
0.865 = 1 - e^(-t/RC)
e^(-t/RC) = 1 - 0.865
Now, we can solve for t:
e^(-t/RC) = 0.135
To isolate t, we take the natural logarithm of both sides:
-ln(e^(-t/RC)) = ln(0.135)
-t/RC = ln(0.135)
t = -RC * ln(0.135)
Now, we substitute the given values:
R = 125 Ω
C = 85.0 µF = 85.0 * 10^(-6) F
t = -(125 Ω) * (85.0 * 10^(-6) F) * ln(0.135)
Calculating this expression will give us the time it takes for the capacitor to rise to 0.865V0, which is approximately two time constants.
Please note that the negative sign arises because we are dealing with the charging of the capacitor in this case.