Class Management Help Chapter 21 Begin Date: 8/11/2017 12:00:00 AM -- Due Date:
ID: 1658219 • Letter: C
Question
Class Management Help Chapter 21 Begin Date: 8/11/2017 12:00:00 AM -- Due Date: 9/25/2017 11:59:00 PM End Date: 12/10/2017 12:00:00 AM (13%) Problem 8: Two capacitors of capacitance 3C and 5C (where C = 0.065 F) are connected in series with a resistor of resistance R= 7.5 . Randomized Variables R=7.5 3C 5C C=0.065 F ©theexpert ta.com 50% Part (a) How long will it take the amount of charge in the circuit to drop by 75% in seconds? Grade Summary Deductions 0% Potential 100% t- sinO | cos() | tan() | | (171819 cotan0asin) acos0 atan acotanOsinh0 coshtanh) cotanh0 Submissions Attempts remaining: 6 (3% per attempt) detailed view 1 2 3 END Degrees Radians VO BACKSPACE CLEAR Submit Hint I give up!Explanation / Answer
here,
as the capacitors are in series
their equivalent capacitance , C' = 3 * C * 5 * C /( 3C + 5 C)
C' = 1.875 C
R = 7.5 ohm
time constant , T = R*C
T = 7.5 * 1.875 * 0.065 s
T = 0.914 s
a)
Q = Q0 * ( 1 - e^(-t/T) )
(1 - 0.75) * Q0 = Q0 * (1 - e^(-t/0.914))
solving for t
t = 0.26 s
time taken is 0.26 s