9-Volt battery is used to charge a 375 mF capacitor over a 147 Ù resistance.
How many seconds will it take for the capacitor to reach 75% of its maximum charge?
Hint: The equation below represents the charge of a charging capacitor as a function of time
Q max= CV
once you find Q max we can equate it to the formula Q=Q max (1-e ^-t/RC)
what do you do with the 75% do you take 75% of the Q max????????/// confused
yes, Q= .75CV, then use the exponential form to find t.
To determine how many seconds it will take for the capacitor to reach 75% of its maximum charge, we can use the equation Q = Qmax(1 - e^(-t/RC)), where Q is the charge on the capacitor, Qmax is the maximum charge, t is time in seconds, R is the resistance, and C is the capacitance.
To find Qmax, we can use the formula Qmax = CV, where C is the capacitance and V is the voltage. In this case, the voltage is 9V and the capacitance is 375 mF (or 375 * 10^-6 F).
So, Qmax = (375 * 10^-6 F)(9V) = 3.375 * 10^-3 C.
Now, we can plug in the values into the equation Q = Qmax(1 - e^(-t/RC)) and solve for t.
0.75 * Qmax = Qmax(1 - e^(-t/(147 Ω * 375 * 10^-6 F)))
0.75 = 1 - e^(-t/(147 Ω * 375 * 10^-6 F))
Taking the natural logarithm (ln) of both sides:
ln(0.75) = -t/(147 Ω * 375 * 10^-6 F)
t/(147 Ω * 375 * 10^-6 F) = -ln(0.75)
Multiplying both sides by (147 Ω * 375 * 10^-6 F):
t = -ln(0.75) * (147 Ω * 375 * 10^-6 F)
Using this equation, you can calculate the time required for the capacitor to reach 75% of its maximum charge.