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4 years ago

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Two capacitors give an equivalent capacitance of 9.42 pF when connected in parallel and an equivalent capacitance of 1.68 pF whe

n connected in series. What is the capacitance of each capacitor?

1 answer:

gavmur [86]4 years ago

5 0

Answer:

$C_1=7.23pF\ and\ C_2=2.19pF$

Explanation:

Let the two capacitance are $C_1\ and\ C_2$

It is given that when capacitors are connected in parallel their equvilaent capacitance is 9.42 pF

So $C_1+ C_2=9.2$ --------EQN 1

And when they are connected in series their equivalent capacitance is 1.68 pF

So $\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{1.68}$

$\frac{C_1+C_2}{C_1C_2}=\frac{1}{1.68}$

$C_1C_2=1.68\times 9.42=15.8256pF$

$C_1-C_2=\sqrt{(C_1+C_2)^2-4C_1C_2}=\sqrt{9.42^2-4\times 15.8256}=5.0432pF$ -----EQN

On solving eqn 1 and eqn 2

$C_1=7.23pF\ and\ C_2=2.19pF$

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