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Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series across a 300 V source. Then
- a)The charge on each capacitor is 4.8 × 10-4 C
- b)The potential difference across C1 is 60 V
- c)The potential difference across C2 is 240 V
- d)The energy stored in the system is 3.6 × 10-2 J
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series ...
In series the equivalent capacitance will be C = 1.6 μF. The charge will be same on each capacitor. Its value will be Q = CV = 1.6 × 10-6 × 300 = 4.8 × 10-4 C. The potential difference across C1 and C2 will be 240 V and 60 V respectively. The energy stored in the system will be 1/2CV2 = 7.2 × 10-2 J.
Most Upvoted Answer
Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series ...
Understanding Capacitors in Series
When capacitors are connected in series, they share the same charge, and the total voltage across the series connection is the sum of the voltages across each capacitor.
Step 1: Calculate Total Capacitance
The formula for total capacitance (C_total) in series is given by:
1/C_total = 1/C1 + 1/C2
For C1 = 2 μF and C2 = 8 μF:
1/C_total = 1/2 + 1/8
Calculating this gives:
1/C_total = 4/8 + 1/8 = 5/8
Thus, C_total = 8/5 = 1.6 μF.
Step 2: Calculate Total Charge (Q)
The charge (Q) on the capacitors can be calculated using:
Q = C_total * V
Where V is the voltage across the source (300 V):
Q = 1.6 μF * 300 V = 4.8 × 10^(-4) C.
Step 3: Voltage Across Each Capacitor
The voltage across each capacitor can be calculated using:
V = Q / C.
- For C1:
V1 = Q / C1 = 4.8 × 10^(-4) C / 2 × 10^(-6) F = 240 V.
- For C2:
V2 = Q / C2 = 4.8 × 10^(-4) C / 8 × 10^(-6) F = 60 V.
Step 4: Energy Stored in the System
The energy (U) stored in the capacitors can be calculated using:
U = 0.5 * C_total * V^2.
U = 0.5 * 1.6 × 10^(-6) F * (300 V)^2 = 0.5 * 1.6 × 10^(-6) * 90000 = 0.072 J.
However, each calculation confirms that the charge on each capacitor is indeed 4.8 × 10^(-4) C, validating option 'A' as correct.
When capacitors are connected in series, they share the same charge, and the total voltage across the series connection is the sum of the voltages across each capacitor.
Step 1: Calculate Total Capacitance
The formula for total capacitance (C_total) in series is given by:
1/C_total = 1/C1 + 1/C2
For C1 = 2 μF and C2 = 8 μF:
1/C_total = 1/2 + 1/8
Calculating this gives:
1/C_total = 4/8 + 1/8 = 5/8
Thus, C_total = 8/5 = 1.6 μF.
Step 2: Calculate Total Charge (Q)
The charge (Q) on the capacitors can be calculated using:
Q = C_total * V
Where V is the voltage across the source (300 V):
Q = 1.6 μF * 300 V = 4.8 × 10^(-4) C.
Step 3: Voltage Across Each Capacitor
The voltage across each capacitor can be calculated using:
V = Q / C.
- For C1:
V1 = Q / C1 = 4.8 × 10^(-4) C / 2 × 10^(-6) F = 240 V.
- For C2:
V2 = Q / C2 = 4.8 × 10^(-4) C / 8 × 10^(-6) F = 60 V.
Step 4: Energy Stored in the System
The energy (U) stored in the capacitors can be calculated using:
U = 0.5 * C_total * V^2.
U = 0.5 * 1.6 × 10^(-6) F * (300 V)^2 = 0.5 * 1.6 × 10^(-6) * 90000 = 0.072 J.
However, each calculation confirms that the charge on each capacitor is indeed 4.8 × 10^(-4) C, validating option 'A' as correct.
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Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series across a 300 V source. Thena)The charge on each capacitor is 4.8 × 10-4 Cb)The potential difference across C1 is 60 Vc)The potential difference across C2 is 240 Vd)The energy stored in the system is 3.6 × 10-2 JCorrect answer is option 'A'. Can you explain this answer?
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Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series across a 300 V source. Thena)The charge on each capacitor is 4.8 × 10-4 Cb)The potential difference across C1 is 60 Vc)The potential difference across C2 is 240 Vd)The energy stored in the system is 3.6 × 10-2 JCorrect answer is option 'A'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series across a 300 V source. Thena)The charge on each capacitor is 4.8 × 10-4 Cb)The potential difference across C1 is 60 Vc)The potential difference across C2 is 240 Vd)The energy stored in the system is 3.6 × 10-2 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series across a 300 V source. Thena)The charge on each capacitor is 4.8 × 10-4 Cb)The potential difference across C1 is 60 Vc)The potential difference across C2 is 240 Vd)The energy stored in the system is 3.6 × 10-2 JCorrect answer is option 'A'. Can you explain this answer?.
Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series across a 300 V source. Thena)The charge on each capacitor is 4.8 × 10-4 Cb)The potential difference across C1 is 60 Vc)The potential difference across C2 is 240 Vd)The energy stored in the system is 3.6 × 10-2 JCorrect answer is option 'A'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series across a 300 V source. Thena)The charge on each capacitor is 4.8 × 10-4 Cb)The potential difference across C1 is 60 Vc)The potential difference across C2 is 240 Vd)The energy stored in the system is 3.6 × 10-2 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two capacitors, C1 = 2 μF and C2 = 8 μF are connected in series across a 300 V source. Thena)The charge on each capacitor is 4.8 × 10-4 Cb)The potential difference across C1 is 60 Vc)The potential difference across C2 is 240 Vd)The energy stored in the system is 3.6 × 10-2 JCorrect answer is option 'A'. Can you explain this answer?.
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