Discussion
Two capacitors of 3 \( \mu \)F and 6 \( \mu \)F in series give a combined capacitance of
Answer: C
Correct answer: 2 \( \mu \)F
In series the reciprocals add: \[ \frac{1}{C} = \frac{1}{3}+\frac{1}{6} = \frac{1}{2}, \qquad C = 2\ \mu\mathrm{F}. \] The result is smaller than either capacitor, which is the check to apply - series capacitance behaves like parallel resistance, and adding them directly gives the 9 offered.
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