You have three capacitors, each of 2 μC. In which of the following combinations of the three capacitors, the resultant capacitance is 6μC?
Correct answer: D. All three capacitors in parallel
- A. All three capacitors in series
- B. Two capacitors are in series, one in parallel
- C. Two capacitors are in parallel, one in series
- D. All three capacitors in parallel
Explanation
For capacitors in parallel, the total capacitance is calculated through;Ct = C1 + C2 + C3Ct = 6μCThis gives the highest net capacitance among the given options. A. For capacitors in series, the total capacitance is calculated through;Ct = 1/C1 + 1/C2 + 1/C3Ct = 1/2μC + 1/2μC + 1/2μCCt = 9/6μC = 1.5μCThe net capacitance is lower than the individual capacitances.B. For two capacitances in series;Ct = 1/C1 + 1/C2 -[1]For one capacitor in parallel;Ct = C3 -[2]Adding [1] and [2],Ct = 1/C1 + 1/C2 + C3Ct = 1/2μC + 1/2μC + 2μCCt= 3μCC. For two capacitors in parallel;Ct = C1 + C2 -[1]For one capacitor in series;Ct = 1/C3 -[2]Adding [1] and [2],Ct = C1 + C2 + 1/C3Ct = 2μC + 2μC + 1/2μCCt = 4μC + 1/2μCCt = 9/2μCCt = 4.5μC
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