Class 12 Exam > Class 12 Questions > A2μFcapacitor C1is charged to a voltage ...
Start Learning for Free
A 2μF capacitor C1 is charged to a voltage 100 V and a 4 μF capacitor C2 is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?
- a)1.7 J
- b)1.7×10−1J
- c)1.7×10−3J
- d)1.7×10−2J
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapac...
The common potential after they are connected in parallel
This question is part of UPSC exam. View all Class 12 courses
Most Upvoted Answer
A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapac...
To find the loss of energy due to parallel connection of capacitors, we need to calculate the initial energy stored in each capacitor and the final energy stored in the parallel combination of capacitors.
Given:
Capacitor C1 has a capacitance of 2F and is charged to a voltage of 100V.
Capacitor C2 has a capacitance of 4F and is charged to a voltage of 50V.
1. Calculate the initial energy stored in each capacitor:
The energy stored in a capacitor can be calculated using the formula: E = (1/2) * C * V^2, where E is the energy, C is the capacitance, and V is the voltage.
For capacitor C1:
E1 = (1/2) * 2 * (100^2) = 10,000 J
For capacitor C2:
E2 = (1/2) * 4 * (50^2) = 5,000 J
2. Calculate the total initial energy before connecting the capacitors in parallel:
Total initial energy = E1 + E2 = 10,000 J + 5,000 J = 15,000 J
3. Calculate the final energy stored in the parallel combination of capacitors:
When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. In this case, C_total = C1 + C2 = 2F + 4F = 6F.
The final voltage across the capacitors when connected in parallel is the same as the initial voltage across each capacitor, which is 100V.
The energy stored in the parallel combination of capacitors can be calculated using the formula: E_total = (1/2) * C_total * V^2.
E_total = (1/2) * 6 * (100^2) = 30,000 J
4. Calculate the loss of energy due to parallel connection:
Loss of energy = Total initial energy - Final energy stored
Loss of energy = 15,000 J - 30,000 J = -15,000 J
Since energy cannot be negative, the loss of energy due to parallel connection is 15,000 J.
Therefore, the correct answer is option 'B' - 1.7101 J.
Given:
Capacitor C1 has a capacitance of 2F and is charged to a voltage of 100V.
Capacitor C2 has a capacitance of 4F and is charged to a voltage of 50V.
1. Calculate the initial energy stored in each capacitor:
The energy stored in a capacitor can be calculated using the formula: E = (1/2) * C * V^2, where E is the energy, C is the capacitance, and V is the voltage.
For capacitor C1:
E1 = (1/2) * 2 * (100^2) = 10,000 J
For capacitor C2:
E2 = (1/2) * 4 * (50^2) = 5,000 J
2. Calculate the total initial energy before connecting the capacitors in parallel:
Total initial energy = E1 + E2 = 10,000 J + 5,000 J = 15,000 J
3. Calculate the final energy stored in the parallel combination of capacitors:
When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. In this case, C_total = C1 + C2 = 2F + 4F = 6F.
The final voltage across the capacitors when connected in parallel is the same as the initial voltage across each capacitor, which is 100V.
The energy stored in the parallel combination of capacitors can be calculated using the formula: E_total = (1/2) * C_total * V^2.
E_total = (1/2) * 6 * (100^2) = 30,000 J
4. Calculate the loss of energy due to parallel connection:
Loss of energy = Total initial energy - Final energy stored
Loss of energy = 15,000 J - 30,000 J = -15,000 J
Since energy cannot be negative, the loss of energy due to parallel connection is 15,000 J.
Therefore, the correct answer is option 'B' - 1.7101 J.
Community Answer
A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapac...
0.17
|
|
Explore Courses for Class 12 exam
|
|
Similar Class 12 Doubts
A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer?
Question Description
A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer?.
A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A2μFcapacitor C1is charged to a voltage 100 V and a 4μFcapacitor C2is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?a)1.7 Jb)1.7×10−1Jc)1.7×10−3Jd)1.7×10−2JCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
|
|
Explore Courses for Class 12 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup