Question

A 0.754F capacitor is charged to 50 V . It is then connected in series with a 558 resistor and a 100 resistor and allowed to
Submitted Answers ANSWER 1: Deduction: -3% 1.1715-10-4 J ANSWER 2: Deduction: -3% 33.3 J ANSWER 3: Deduction: -3% 33.3-10-6J
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Here we have given that,

V = 50 V

C = 0.75uF

R1 = 55 ohm

R2 = 100 ohm

Since we know that,

Energy stored in the capacitor is given as,
E = ½CV² = ½QV J

So that here, Ratio of energy dissipated by resistor compared to total resistance will be
R1/(R1+R2)

So that the portion of energy dissipated by R1 is
[R1/(R1+R2)] * ½CV²

E(55 ohm)= [55/(55+100)] (0.5) (0.75×10^-6) (50 )²

= (3.3266129) × 10^-4 J

Add a comment
Know the answer?
Add Answer to:
A 0.754F capacitor is charged to 50 V . It is then connected in series with...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT