A charged capacitor, C = 40 micro F, is connected to a resistor, 120 ohms. How much time will elapse until the voltage falls to 25% of its original (max) value?
im not even sure what equation to use for this, so if someone could help me to figure that out it wou
ld be greatly appreciated!
R = 120 Ohms.
C = 40 uF = 40*10^-6 Farads.
RC = 120 * 40*10^-6 = 4.8*10^-3 seconds
t/RC = t/4.8*10^-3 = 208.3t
1/e^208.3t = 0.25
e^208.3t = 4
208.3t*Ln e = Ln 4
208.3t = 1.386
t = 0.00666 s. = 6.66 Milliseconds
To solve this problem, we can use the equation for the voltage across a charging or discharging capacitor in an RC circuit:
V = V0 * e^(-t/RC)
Where:
- V is the voltage across the capacitor at any given time
- V0 is the initial voltage (max value)
- t is the time that has elapsed
- R is the resistance of the resistor
- C is the capacitance of the capacitor
In this case, we are given:
- V0 = maximum voltage = initial voltage
- C = 40 micro F (40 * 10^(-6) F)
- R = 120 ohms
We want to find the time it takes for the voltage to fall to 25% of its original value, so V = 0.25 * V0.
0.25 * V0 = V0 * e^(-t/RC) (substitute V = 0.25 * V0)
Now, let's solve for t:
0.25 = e^(-t/RC)
To isolate t and eliminate the exponential, we can take the natural logarithm (ln) of both sides:
ln(0.25) = ln(e^(-t/RC))
Using the property of logarithms (log_b(x^a) = a * log_b(x)), this simplifies to:
ln(0.25) = (-t/RC) * ln(e)
Since ln(e) = 1:
ln(0.25) = -t/RC
Now, let's solve for t by rearranging the equation:
t = -RC * ln(0.25)
Substituting the given values:
t = -(120 ohms) * (40 * 10^(-6) F) * ln(0.25)
Using a calculator to evaluate the natural logarithm:
t ≈ -(120) * (40 * 10^(-6)) * (-1.3863)
Simplifying the expression:
t ≈ 0.0052596 s
Therefore, it takes approximately 0.0052596 seconds (or 5.2596 milliseconds) for the voltage across the capacitor to fall to 25% of its original value.