Question

two capacitor are connected in series across2kv line the energy stored in the system is 2.4j if the same capacitor connected in parallel across the same line the energy stored is 10j find individual capacitors ​

Answer 1

Explanation:

HEY MATE ..

Let the capacitance of two capacitors be C 1and C 2

Line voltage V=4000 voltsParallel connection :Equivalent capacitance of series connection C p =C 1+C 2

Energy stored Ep = 1/2 C p V ^2

∴ 36= 1 / 2 (C 1+C 2)(4000) ^2

⟹ C 1 +C 2 =4.5μF

=4.5μFWe get C 2 =4.5μF−C 1

Series connection :

Series connection :Equivalent capacitance of series connection C s =

C1 .C2 /C1 + C2 =C1.C2 /4.5 μF

Energy stored E s = 1/2 C s.V ^2

∴ 8= 1/2

4.5 / ( C1 . C 2 )(4000) ^2

⟹ C 1 . C 2 =4.5μF

=4.5μFOr C 1 (4.5−C 1)=4.5

)=4.5Or C 1/2−4.5C 1 +4.5=0

+4.5=0Solving we get C 1 =1.5μF

=1.5μF⟹ C 2 =4.5−1.5=3.0μF