JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 14)
Effective capacitance of parallel combination
of two capacitors C1 and C2 is 10 μF. When
these capacitors are individually connected to
a voltage source of 1V, the energy stored in the
capacitor C2 is 4 times that of C1. If these
capacitors are connected in series, their
effective capacitance will be :
4.2 μF
8.4 μF
1.6 μF
3.2 μF
Explanation
C1
+ C2
= 10 ......(1)
$${1 \over 2}{C_2}{V^2} = 4 \times {1 \over 2}{C_1}{V^2}$$
$$ \Rightarrow $$ C2 = 4C1 .....(2)
From (1) ans (2),
C1 = 2 and C2 = 8
For series combination
Ceq = $${{{C_1}{C_2}} \over {{C_1} + {C_2}}}$$ = $${{8 \times 2} \over {8 + 2}}$$ = 1.6 $$\mu $$F
$${1 \over 2}{C_2}{V^2} = 4 \times {1 \over 2}{C_1}{V^2}$$
$$ \Rightarrow $$ C2 = 4C1 .....(2)
From (1) ans (2),
C1 = 2 and C2 = 8
For series combination
Ceq = $${{{C_1}{C_2}} \over {{C_1} + {C_2}}}$$ = $${{8 \times 2} \over {8 + 2}}$$ = 1.6 $$\mu $$F
Comments (0)
