Question 6 please Thank youu! A capacitor with C = 0.29 mu F is initially to a p

Question

Question 6 please
Thank youu!

A capacitor with C = 0.29 mu F is initially to a potential of 6.60 V The capacitor is then connected across a resistor and allowed to discharge. After a time of 3.00 times 10^-3 s, the potential across the capacitor has dropped to 1.20 V. Calculate the value of the resistance. a) Two capacitors (C_1 = 4.3 mu F, C_2 = 16.4 mu F) are charged individually to (V_1 = 18.6 V, V_2 = 4.8 V). The two capacitors are then connected together in parallel with the positive plates together and the negative places together Calculate the final potential difference across the plates of the capacitors of the capacitors once they are connected. b) Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together. c) By how much (absolute value) is the total stored energy reduced when the two capacitors are connected?

Explanation / Answer

6) C1 = 4.3 uF

C2 = 16.4 uF

V1 = 18.6 V

V2 = 4.8 V

Q1 ( initial charge on C1) = C1*V1 = 4.3*10-6*18.6 = 7.9*10-5 C

Q2 = C2*V2 = 16.4*10-6*4.8 = 7.87*10-5C

a) Let the final potential across both capacitors will be V'.

now Q1' = C1*V'

Q2' = C2*V'

from conservation of charges :

Q1+Q2 = Q1' + Q2'

7.9*10-5 C + 7.87*10-5C = C1*V' + C2*V' = (C1+C2)*V'

1.57*10-4 = ( 4.3 + 16.4)*10-6 *V'

V' ( potential difference when capacitors are connected in parallel) = 7.58 V

-----------------------------------

b) Q1' = C1*V = 4.3*10-6*7.58 = 3.25*10-5 C

But initially Q1 = 7.9*10-5 C

so charge flows will be = Q1 - Q1' = 4.65*10-5 C

-----------------------------------------------------

c) initial energy ,E1 = ( 1/2)*C1*V12 + (1/2)*C2*V22 = 9.32*10-4 J

final energy ,E1' = (1/2)*( C1+C2)*V'2 = 5.94*10-4J

energy reduced = E1 - E1' = 3.38*10-4J

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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