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Calculate the equivalent capacitance for three capacitors in series with values 6 μF, 4 μF, and 3 μF.

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Answer

C) 4/3 μF\text{C) } 4/3 \, \mu\text{F}

Step-by-step solution

6 steps

The capacitors are connected in series. For capacitors in series, the reciprocal of the equivalent capacitance (CeqC_{eq}) is the sum of the reciprocals of individual capacitances.

  1. Identify the given capacitance values.

    C1=6 μFC_1 = 6 \, \mu\text{F}
    C2=4 μFC_2 = 4 \, \mu\text{F}
    C3=3 μFC_3 = 3 \, \mu\text{F}

  2. Apply the formula for equivalent capacitance in series.

    1Ceq=1C1+1C2+1C3\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

    Substitute the given values:

    1Ceq=16 μF+14 μF+13 μF\frac{1}{C_{eq}} = \frac{1}{6 \, \mu\text{F}} + \frac{1}{4 \, \mu\text{F}} + \frac{1}{3 \, \mu\text{F}}

  3. Find a common denominator for the fractions, which is 12.

    1Ceq=212 μF+312 μF+412 μF\frac{1}{C_{eq}} = \frac{2}{12 \, \mu\text{F}} + \frac{3}{12 \, \mu\text{F}} + \frac{4}{12 \, \mu\text{F}}

  4. Add the fractions.

    1Ceq=2+3+412 μF=912 μF\frac{1}{C_{eq}} = \frac{2+3+4}{12 \, \mu\text{F}} = \frac{9}{12 \, \mu\text{F}}

  5. Simplify the fraction.

    1Ceq=34 μF\frac{1}{C_{eq}} = \frac{3}{4 \, \mu\text{F}}

  6. Invert the fraction to find CeqC_{eq}.

    Ceq=43 μFC_{eq} = \frac{4}{3} \, \mu\text{F}

    The calculated equivalent capacitance is 43 μF\frac{4}{3} \, \mu\text{F}. This matches option (C).

    The final answer is C) 4/3 μF\text{C) } 4/3 \, \mu\text{F}.

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