Question

in the given network the value of c so that the equivalent capacitance between points a and b is 3×10^-6​

Answer 1

Answer:

Value of capacitance of the capacitor is given as

$C = 3.68 \times 10^{-6} F$

Explanation:

First of all two 2 uF capacitors are in series then it is parallel with 1 uF

so we will have

$C = (\frac{1}{\frac{1}{2} + \frac{1}{2}}) + 1$

$C_1 = 2\mu F$

Now this is in series with 8 uF so we have

$C_2 = \frac{1}{\frac{1}{8} + \frac{1}{2}}$

$C_2 = 1.6 \mu F$

now it is in parallel with 12 uF so we have

$C_3 = 12 \mu + 1.6 \mu$

$C_3 = 13.6 \mu F$

Now we have 2 uF and 6 uF in parallel which is in series with 4 uF

so we have

$C_4 = \frac{1}{\frac{1}{6 + 2} + \frac{1}{4}}$

$C_4 = \frac{8}{3}\mu F$

now these two are in parallel so we have

$C = \frac{8}{3} + 13.6$

$C = 16.27 \mu F$

Now it is in series with C and equivalent is give to us

so we have

$\frac{1}{3\times 10^{-6}} = \frac{1}{C} + \frac{1}{16.27 \times 10^{-6}}$

$C = 3.68 \times 10^{-6} F$

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Topic : equivalent capacitor

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