Please help calculating the time for this capacitor problem? I keep getting 0.46

Question

Please help calculating the time for this capacitor problem? I keep getting 0.4654 seconds, but it says it's wrong :( please help!

(20%) Problem 3: Two capacitors of capacitance 3C and 5C (where C = 0.055 F) are connected in series with a resistor of resistance R = 7.5 Randomized Variables R=7.5 3C 5C C=0.055 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 0% 100% cosO cotan0 asin) acos0 Submissions Attempts remaining: 18 sinO HOME % per attempt) detailed view sinhO atan)acotan) si coshO tanhO cotanh0 0% 0% 0 END Degrees Radians 0 DELCLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 3 Feedback: 0%-deduction per feedback. Submission History Hints Answer t = 0.4654 Feedback Note: Feedback not accessed Totals 0% 0% 0% 0%

Explanation / Answer

Equivalent Capacitance = 3C*5C/(3C+5C)

= 15/8 C = 15/8*0.055 = 0.1031 F

Now by tge equation of charging of capacitor,

q/Q0 = 1-e^(-t/RC)

0.75 = 1- e^(-t/(75*0.1031))

e^(-t/(75*0.1031)) = 1-0.75 = 0.25

- t/(75*0.1031) = ln 0.25

t = - 75*0.1031 ln 0.25

= 10.72 s answer

  

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