Two capacitors of capacitance (6 0.09) F abd (3 0.09) F are connected in series. The…
Physics · JEE Main · NTA Exams — Electrostatics
Two capacitors of capacitance \((6 \pm 0.09) \mu \mathrm{F} \text { abd }(3 \pm 0.09) \mu \mathrm{F}\) are connected in series. The equivalent capacitance C with error is
- \((2 \pm 0.05) \mu \mathrm{F}\)
- \((2 \pm 0.18) \mu F\)
- \((2 \pm 0.20) \mu \mathrm{F}\)
- \((2 \pm 0.04) \mu \mathrm{F}\)
Answer
(A) (2 0.05) F
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