Series CombinationEasyQuestion 31522
Question
Two capacitors 4 μF and 12 μF are connected in series. Equivalent capacitance is:
- A16 μF
- B3 μFCorrect
- C8 μF
- D48 μF
Correct answer
3 μF
Explanation
1/C = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, so C = 3 μF.
Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.
1/C = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, so C = 3 μF.