given: CS=3Mf CP=16Mfas CS=[(C1C2)/(C1+C2)]=3CP=C1+C2=16(2)∴from (1), [(C1C2)/(16)]=3C1C2=48from (2), C2=16−C1C2=16−(48/C2)∴C2+(48/C2)=16∴C22+48−16C2=0∴C2=12μF,C2=4μF i.e. capacitance are 4μF & 12μF
NEET UG 2022 — Physics Electromagnetism
The effective capacitances of two capacitors are 3 μF and 16μF, when they are connected in series and parallel respectively. The capacitance of two capacitors are:
Held on 30 Apr 2022 · Verified 9 Jul 2026.
10μF,6μF
8μF,8μF
12μF,4μF
1.2μF,1.8μF
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