(13%) Problem 8: Two capacitors of capacitance 3C and 5C (where C = 0.12 F) are
ID: 1658155 • Letter: #
Question
(13%) Problem 8: Two capacitors of capacitance 3C and 5C (where C = 0.12 F) are connected in series with a resistor of resistance R = 6.5 Randomized Variables R=6.5 3C 5C C=0.12 F ©theexpertta.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 Potential 100% Submissions Attempts remaining: 6 (3% per attempt) detailed view 789 cosO cotanO asinO sin) acos atan) acotan)sinhO cosh)tanh) cotanh( END Degrees O Radians CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback 50% Part (b) If the circuit was charged by a 10.0 V source how much total charge (in C) did both capacitors have in them to begin with?Explanation / Answer
Net capacitance of the circuit is Cnet = (3C*5C)/(3C+5C) = 1.875*C = 1.875*0.12 = 0.225 F
R = 6.5 ohm
Time constant is T = R*Cnet = 6.5*0.225 = 1.463 sec
while discharging
charge on capacitor after t sec is
q = Q*e^(-t/T)
given that q = 0.75*Q
then
0.75*Q = Q*e^(-t/1.463)
0.75 = e^(-t/1.463)
t = 0.42 sec
-----------------------------------------------------
b) Q = Cnet*V = 0.225*10 = 2.25 C
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.