A 9-Volt battery is used to charge a 280 mF capacitor over a 81 Ohms resistance.
How long will it take for the capacitor to reach 75% of its maximum charge?
What formula is relevant?
charging the capacitor formula;
Q=Q max (1-e ^-t/RC)
Q=qmax= 0.75
solve for t
So the Voltage is irrelevant?
Why wouldn't you use this formula?
Q=Qmax*e^(-t/RC)
To calculate the time it takes for a capacitor to reach a certain charge, we can use the formula for the charging of a capacitor through a resistor:
t = RC ln(Vf/Vi)
Where:
t = time taken (in seconds)
R = resistance (in ohms)
C = capacitance (in farads)
ln stands for natural logarithm
Vi = initial voltage (in volts)
Vf = final voltage (in volts)
In this case, the values given are:
R = 81 ohms
C = 280 mF = 280 * 10^-3 F
Vi = 0 volts (since the capacitor is starting from 0 charge)
Vf = 75% of the maximum charge = 0.75 * 9 volts
Plugging these values into the formula, we have:
t = (81 ohms) * (280 * 10^-3 F) * ln(0.75 * 9 volts / 0 volts)
Now we can proceed to solve this equation.