Suppose a capacitor having a capacitance of C fully charged to a voltage V and then discharged through a resistor R. Derive the equation (in terms of C,V,and R) that shows how long a time the capacitor will take to lose 45 % of its charge.
Q = V C
V = i R
i = - dQ/dt
so
i R = Q/C
-dQ/dt = Q/(RC)
-dQ/Q = dt/(RC)
-ln Q = - t/(RC)
Q = k e^[- t/(RC) ] as we all knew
when t = 0 Q = Qi
so
Q = Qi e^[-t/(RC)]
when is Q = .55 Qi ??
.55 = e^[-t/(RC)]
ln .55 = -.5978 = -t/(RC)
so
t = 0.5978 R C
To derive the equation for the time it takes for a capacitor to lose 45% of its charge when discharged through a resistor, we can use the formula for the voltage across a discharging capacitor as a function of time.
The voltage across a discharging capacitor is given by the equation:
V(t) = V(0) * exp(-t / (RC))
Where:
- V(t) is the voltage across the capacitor at time t.
- V(0) is the initial voltage across the capacitor (fully charged voltage).
- R is the resistance in the circuit.
- C is the capacitance of the capacitor.
- e is the base of the natural logarithm.
We want to find the time it takes for the voltage to decay to 45% of its initial value V(0).
Therefore, we can set V(t) equal to 0.45 * V(0) and solve for t:
0.45 * V(0) = V(0) * exp(-t / (RC))
Dividing both sides by V(0), we get:
0.45 = exp(-t / (RC))
Taking the natural logarithm (ln) of both sides, we have:
ln(0.45) = ln(exp(-t / (RC)))
Using the property ln(exp(x)) = x, we simplify to:
ln(0.45) = -t / (RC)
Multiplying both sides by -(RC), we obtain:
-t / C = ln(0.45) * R
Finally, multiplying both sides by -1 and rearranging the equation, we get the equation for the time it takes for the capacitor to lose 45% of its charge:
t = -ln(0.45) * R * C
Therefore, the equation in terms of C, V, and R that shows how long a time the capacitor will take to lose 45% of its charge is:
t = -ln(0.45) * R * C