Capacitor C 3 in the circuit is a variable capacitor (its capacitance can be varied). Graph is plotted between potential difference V 1 (across capacitor C 1 ) versus C 3 . Electric potential V 1 approaches on asymptote of 10 V as C 3 → ∞ .
(i) EMF of the battery is equal to:
Text Solution
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(i)
(ii)
(iii)
Sol. When C 3 = ∞ , there will be no charge on C 2
As V 1 = 10 V, therefore V = 10 V
From graph when C 3 = 10 µF, V 1 = 6 V
Charge on C 1 = Charge on C 2 + Charge on C 3
6C 1 = 4C 2 + 40 µC ....
Also when C 3 = 6 µF, V 1 = 5V
Again using charge equation
5C 1 = 5C 2 + 30 µC....
Solving and
C 1 = 8 µ F
C 2 = 2 µ F.
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