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The equivalent capacitance of the arrangement shown in figure is
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The correct answer is:
$20 \mu \mathrm{F}$
We know that the, circuit can be modified as
$\therefore C_{\text {net }}=15+5=20 \mu \mathrm{F}$
$\therefore C_{\text {net }}=15+5=20 \mu \mathrm{F}$
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