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QJEEM20Ph3742: JEE Main 2020 Physics - Class 12 JEEM20Ph3742 Solution: A 10 F capacitor is fully charged to a potential difference of…
A 10 F capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :
Multiple Choice Options
- A 20 F
- B 15 F
- C 10 F
- D 30 F
Step-by-Step Verified Solution
Initially,
Charge on capacitor 10 μF
Q = CV = (10 μF) (50V)
Q = 500 μC
Final Charge on 10 μF capacitor
Q = CV = (10 μF) (20V)
Q = 200 μC
From charge conservation,
Charge on unknown capacitor
Q = 500 μC – 200 μC = 300 μC
Capacitance (C) = = = 15 F
Charge on capacitor 10 μF
Q = CV = (10 μF) (50V)
Q = 500 μC
Final Charge on 10 μF capacitor
Q = CV = (10 μF) (20V)
Q = 200 μC
From charge conservation,
Charge on unknown capacitor
Q = 500 μC – 200 μC = 300 μC
Capacitance (C) = = = 15 F