Given:
\(C1 = 1 \mu F\)
\(C2 = 2 \mu F\)
The equivalent capacitance for capacitors connected in series is given by:
\[ C_{eq} = \frac{C1 \times C2}{C1 + C2} \]
Substitute the given values:
\[ C_{eq} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \mu F = 0.67 \mu F \]
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