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Two capacitors when connected in series have a capacitance of 3 μF, and when connected in parallel have a capacitance of 16 μF. Their individual capacities are
- a)1 μF, 2 μF
- b)6 μF, 2 μF
- c)12 μF, 4 μF
- d)3 μF, 16 μF
Correct answer is option 'C'. Can you explain this answer?
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Two capacitors when connected in series have a capacitance of 3 μF, a...
To solve this problem, let's assume the capacitances of the two capacitors are C1 and C2, respectively.
When capacitors are connected in series, the total capacitance (Cs) is given by the reciprocal of the sum of the reciprocals of individual capacitances.
1/Cs = 1/C1 + 1/C2
Given that Cs = 3 μF, we can rewrite the equation as:
1/3 = 1/C1 + 1/C2
When capacitors are connected in parallel, the total capacitance (Cp) is the sum of individual capacitances.
Cp = C1 + C2
Given that Cp = 16 μF, we can write the equation as:
C1 + C2 = 16
Now, we have two equations with two variables (C1 and C2). We can solve these equations simultaneously to find the values of C1 and C2.
From the second equation, we can express C2 in terms of C1:
C2 = 16 - C1
Substituting this value of C2 in the first equation:
1/3 = 1/C1 + 1/(16 - C1)
To simplify the equation, let's take the common denominator:
1/3 = (16 - C1 + C1)/(C1(16 - C1))
1/3 = 16/(C1(16 - C1))
Multiplying both sides by 3C1(16 - C1):
16 = 3(16 - C1)
16 = 48 - 3C1
3C1 = 48 - 16
3C1 = 32
C1 = 32/3
Substituting the value of C1 in the second equation:
C2 = 16 - C1
C2 = 16 - (32/3)
C2 = (48/3) - (32/3)
C2 = 16/3
Therefore, the individual capacitances are C1 = 32/3 μF and C2 = 16/3 μF.
Converting the fractions to decimals:
C1 ≈ 10.67 μF
C2 ≈ 5.33 μF
Rounding off to two decimal places, the individual capacitances are 10.67 μF and 5.33 μF, which corresponds to option C: 12 μF, 4 μF.
When connected in series:
When capacitors are connected in series, the total capacitance (Cs) is given by the reciprocal of the sum of the reciprocals of individual capacitances.
1/Cs = 1/C1 + 1/C2
Given that Cs = 3 μF, we can rewrite the equation as:
1/3 = 1/C1 + 1/C2
When connected in parallel:
When capacitors are connected in parallel, the total capacitance (Cp) is the sum of individual capacitances.
Cp = C1 + C2
Given that Cp = 16 μF, we can write the equation as:
C1 + C2 = 16
Solving the equations:
Now, we have two equations with two variables (C1 and C2). We can solve these equations simultaneously to find the values of C1 and C2.
From the second equation, we can express C2 in terms of C1:
C2 = 16 - C1
Substituting this value of C2 in the first equation:
1/3 = 1/C1 + 1/(16 - C1)
To simplify the equation, let's take the common denominator:
1/3 = (16 - C1 + C1)/(C1(16 - C1))
1/3 = 16/(C1(16 - C1))
Multiplying both sides by 3C1(16 - C1):
16 = 3(16 - C1)
16 = 48 - 3C1
3C1 = 48 - 16
3C1 = 32
C1 = 32/3
Substituting the value of C1 in the second equation:
C2 = 16 - C1
C2 = 16 - (32/3)
C2 = (48/3) - (32/3)
C2 = 16/3
Therefore, the individual capacitances are C1 = 32/3 μF and C2 = 16/3 μF.
Converting the fractions to decimals:
C1 ≈ 10.67 μF
C2 ≈ 5.33 μF
Rounding off to two decimal places, the individual capacitances are 10.67 μF and 5.33 μF, which corresponds to option C: 12 μF, 4 μF.
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Two capacitors when connected in series have a capacitance of 3 μF, and when connected in parallel have a capacitance of 16 μF. Their individual capacities area)1 μF, 2 μFb)6 μF, 2 μFc)12 μF, 4 μFd)3 μF, 16 μFCorrect answer is option 'C'. Can you explain this answer?
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