JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 8)
A 10 $$\mu $$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 :
20 $$\mu $$F
15 $$\mu $$F
10 $$\mu $$F
30 $$\mu $$F
Explanation
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
$$ \Rightarrow $$ Capacitance (C) = $${Q \over V}$$ = $${{300} \over {20}}$$ = 15 $$\mu $$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
$$ \Rightarrow $$ Capacitance (C) = $${Q \over V}$$ = $${{300} \over {20}}$$ = 15 $$\mu $$F
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