Question

when two capacitatora of capacitance C1 and C2 are connected in series, the net capacitance is 3 muF, when connected in parallel, the net capacitance is 16 muF. Calculate the values of C1 and C2​

Answer 1

Explanation:

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Answer 2

Answer:

The values of C1 and C2​ are 12uF and 4uF respectively.

Explanation:

The reciprocal of the equivalent capacitance when numerous capacitors are linked in series is equal to the total of the reciprocals of the individual capacitances. The equivalent capacitance is the total of the individual capacitances when many capacitors are connected in a parallel configuration.

The overall capacitance of a series of capacitors is less than the sum of the individual capacitances of the capacitors. When two or more capacitors are linked in series, the final result is the equivalent of a single capacitor with the total of the individual capacitors' plate spacings. As we've just seen, while all other things remain constant, capacitance decreases as plate gap increases.

Capacitance connected in series,

$\frac{1}{C_{s} }=\frac{1}{C_{1} }+\frac{1}{C_{2} } +\frac{1}{C_{3} } +.......... \frac{1}{C_{n} }$

Capacitance connected in parallel,

$C_{p}=C_{1} +C_{2} +C_{3} +................+C_{n}$

Given:

Two capacitators of capacitance C1 and C2 are connected in series, the net capacitance is 3uF.

$\frac{1}{C_{1} }+\frac{1}{C_{2} } =\frac{1}{3}$

$\frac{C_{1}C_{2} }{C_{1}+C_{2}}=3$...........(1)

When connected in parallel, the net capacitance is 16uF.

${C_{1}+{C_{2} =16$ ........ (2)

Substitute equation(2) in equation (1)

$\frac{C_{1}C_{2} }{16}=3\\\\C_{1}C_{2}=48$

$(C_{1}+C_{2})^{2}=C_{1}^{2}+C_{2}^{2}+2C_{1}C_{2}\\16^{2} =C_{1}^{2}+C_{2}^{2}+2\times48\\C_{1}^{2}+C_{2}^{2}=256-96\\C_{1}^{2}+C_{2}^{2}=160$

Now,

$(C_{1}-C_{2})^{2}=C_{1}^{2}+C_{2}^{2}-2C_{1}C_{2}\\(C_{1}-C_{2})^{2}=160-2\times48\\(C_{1}-C_{2})^{2}=160-96\\(C_{1}-C_{2})^{2}=64$

$(C_{1}-C_{2})=8$...….. (3)

Add equation (2) and (3)

$\\(C_{1}+C_{2})+(C_{1}-C_{2})=16+8\\2C_{1} =24$

$C_{1} =12uF$.....(4)

Substitute equation (4) in equation (2), we get

${C_{1}+{C_{2} =16uF$

$12+C_{2} =16uF\\C_{2} =16-12uF$

$C_{2} =4uF$

Hence, the values of C1 and C2​ are 12uF and 4uF respectively.

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