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A capacitance of 2uf is required in an electrical circuit across a potential difference of 1.0 kV.Alarge number of 1uf capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is: A.32 B.2 C.16 D.24?
Most Upvoted Answer
A capacitance of 2uf is required in an electrical circuit across a pot...
Solution:
Given: Capacitance required= 2 µF, Potential difference= 1 kV, Capacitors available= 1 µF, Potential difference of each capacitor= 300 V.
Let's assume that x number of capacitors are required to achieve the required capacitance of 2 µF.
Thus, the formula of capacitors in series is used to find the equivalent capacitance:
1/Ceq = 1/C1 + 1/C2 + 1/C3 + ... + 1/Cn
where Ceq is the equivalent capacitance, C1, C2, C3, ..., Cn are the capacitances of the capacitors and n is the number of capacitors in the circuit.
Now, putting the given values in the above formula, we get:
1/Ceq = 1/1 + 1/1 + 1/1 + ... + 1/1 (n times)
1/Ceq = n/1
Ceq = 1/n
Since, we need to find the minimum number of capacitors required, we can assume that all the capacitors are connected in series, as it will require the minimum number of capacitors.
Thus, putting the values of Ceq and C in the above formula, we get:
1/2 = n/1
n = 2
Therefore, the minimum number of capacitors required to achieve the required capacitance of 2 µF is 2.
Answer: B.2
Explanation:
To achieve the required capacitance of 2 µF across a potential difference of 1 kV, we need to use a combination of capacitors such that the equivalent capacitance is 2 µF and can withstand a potential difference of 1 kV.
Since we have 1 µF capacitors with a potential difference rating of 300 V, we need to connect them in series to increase the equivalent capacitance and withstand the potential difference of 1 kV.
Using the formula of capacitors in series, we can find the equivalent capacitance of the combination of capacitors.
Assuming that all the capacitors are connected in series, we can find the minimum number of capacitors required by putting the values of equivalent capacitance and individual capacitance in the formula.
Thus, the minimum number of capacitors required is 2.
Given: Capacitance required= 2 µF, Potential difference= 1 kV, Capacitors available= 1 µF, Potential difference of each capacitor= 300 V.
Let's assume that x number of capacitors are required to achieve the required capacitance of 2 µF.
Thus, the formula of capacitors in series is used to find the equivalent capacitance:
1/Ceq = 1/C1 + 1/C2 + 1/C3 + ... + 1/Cn
where Ceq is the equivalent capacitance, C1, C2, C3, ..., Cn are the capacitances of the capacitors and n is the number of capacitors in the circuit.
Now, putting the given values in the above formula, we get:
1/Ceq = 1/1 + 1/1 + 1/1 + ... + 1/1 (n times)
1/Ceq = n/1
Ceq = 1/n
Since, we need to find the minimum number of capacitors required, we can assume that all the capacitors are connected in series, as it will require the minimum number of capacitors.
Thus, putting the values of Ceq and C in the above formula, we get:
1/2 = n/1
n = 2
Therefore, the minimum number of capacitors required to achieve the required capacitance of 2 µF is 2.
Answer: B.2
Explanation:
To achieve the required capacitance of 2 µF across a potential difference of 1 kV, we need to use a combination of capacitors such that the equivalent capacitance is 2 µF and can withstand a potential difference of 1 kV.
Since we have 1 µF capacitors with a potential difference rating of 300 V, we need to connect them in series to increase the equivalent capacitance and withstand the potential difference of 1 kV.
Using the formula of capacitors in series, we can find the equivalent capacitance of the combination of capacitors.
Assuming that all the capacitors are connected in series, we can find the minimum number of capacitors required by putting the values of equivalent capacitance and individual capacitance in the formula.
Thus, the minimum number of capacitors required is 2.
Community Answer
A capacitance of 2uf is required in an electrical circuit across a pot...
16 capacitors-
4 parallel connections with 4 capacitors each.
Each capacitor of 1 uf has 250 volts across it.
All conditions are met.
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A capacitance of 2uf is required in an electrical circuit across a potential difference of 1.0 kV.Alarge number of 1uf capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is: A.32 B.2 C.16 D.24?
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A capacitance of 2uf is required in an electrical circuit across a potential difference of 1.0 kV.Alarge number of 1uf capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is: A.32 B.2 C.16 D.24? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A capacitance of 2uf is required in an electrical circuit across a potential difference of 1.0 kV.Alarge number of 1uf capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is: A.32 B.2 C.16 D.24? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A capacitance of 2uf is required in an electrical circuit across a potential difference of 1.0 kV.Alarge number of 1uf capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is: A.32 B.2 C.16 D.24?.
A capacitance of 2uf is required in an electrical circuit across a potential difference of 1.0 kV.Alarge number of 1uf capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is: A.32 B.2 C.16 D.24? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A capacitance of 2uf is required in an electrical circuit across a potential difference of 1.0 kV.Alarge number of 1uf capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is: A.32 B.2 C.16 D.24? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A capacitance of 2uf is required in an electrical circuit across a potential difference of 1.0 kV.Alarge number of 1uf capacitors are available which can withstand a potential difference of not more than 300V. The minimum number of capacitors required to achieve this is: A.32 B.2 C.16 D.24?.
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