Asked by nat
Two capacitors, C1=25.0x10^-6 and C2=5.0x10^-6 are connected in parallel and charged with a 100-V power supply.
a)calculate the total energy stored in the 2 capacitors
b) what potential difference would be required across the same 2 capacitors connected in series in order that the combination store the same energy as in a)?
ENERGY STORED=1/2C*DELTA V^2
1/2(25.0e^-6 + 5.0e^-6)(100)^2
=0.15KJ OR 150J, IS THIS RIGHT? IF NOT HOW DO I GET THE RIGHT ANSWER?
B)1/Ceq= 1/25.0e^-6 + 1/5.0e^-6= 4.16e^-6
DELTA V= SQRT 2x (ENERGY STORED=150J)/
4.16e^-6 = 8492v?????? CAN SOMEONE TELL ME IF IM DOING THIS RIGHT?
a)calculate the total energy stored in the 2 capacitors
b) what potential difference would be required across the same 2 capacitors connected in series in order that the combination store the same energy as in a)?
ENERGY STORED=1/2C*DELTA V^2
1/2(25.0e^-6 + 5.0e^-6)(100)^2
=0.15KJ OR 150J, IS THIS RIGHT? IF NOT HOW DO I GET THE RIGHT ANSWER?
B)1/Ceq= 1/25.0e^-6 + 1/5.0e^-6= 4.16e^-6
DELTA V= SQRT 2x (ENERGY STORED=150J)/
4.16e^-6 = 8492v?????? CAN SOMEONE TELL ME IF IM DOING THIS RIGHT?
Answers
Answered by
drwls
The stored energy that you calculate with that formula is in J, not kJ. The answer to the first part is 0.15 J
For the second part, require that
(1/2)Ceq*V^2 = 0.15 J, and solve for V. You are correct that Ceq = 4.167*10^-6 F
For the second part, require that
(1/2)Ceq*V^2 = 0.15 J, and solve for V. You are correct that Ceq = 4.167*10^-6 F
Answered by
nat
so for part two i get 268V is that correct?
Answered by
drwls
Yes
Answered by
Anonymous
yes
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.