Problem 8: Two capacitors of capacitance 3C and 5C (where C = 0.11 F) are connec

Question

Problem 8: Two capacitors of capacitance 3C and 5C (where C = 0.11 F) are connected in series with a resistor of resistance R = 9.5 . Randomized Variables R=9.5 3C 5C C=0.11 F Otheexpertta.com Part (a) How long will it take the amount of charge in the circuit to drop by 75% in seconds? Numeric :A numeric value is expected and not an expression. 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? Numeric : A numeric value is expected and not an expression. qi

Explanation / Answer

Total capacitance is given for series connection

C = (1/3 + 1/5)-1 = 15/8 x 0.11 F = 0.20625 Farad

Hence time constant RC = 9.5 x 0.20625 = 1.959375 seconds

As q = qoexp(-t/RC)

Hence for q = 0.25qo

t = -RC ln(0.25) = 2.716 seconds

Time required to drop charge by 75% = 2.716 s

(b) as q = CV

C = 0.20625 F

V = 10 V

Hence, charge total in capacitors = 2.0625 C

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.