When three capacitors are connected in series, the equivalent capacitance will be
"C=\\frac{1}{1\/C_1+1\/C_2+1\/C_3}=\\\\\n\\space\\\\\n=\\frac{1}{1\/6+1\/6+1\/12}=2.4\\space\\mu\\text{F}."
Now solve another problem, that is, find what capacitance must be added in series if the total capacitance is 1.2 µF:
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