Each capacitor in figure has a capacitance of 10 µF. The emf of the battery is 100 V. Find the energy stored in each of the four capacitors.
Solution
Capacitors b and c are in parallel; their equivalent capacitance is 20 µF.
Thus, the net capacitance of the circuit is given by
`1/C_"net" = 1/10 + 1/20 + 1/10`
`⇒ 1/C_"net" = (2+1+2)/20 = 5/20`
`⇒ C_"net" = 4 "uF"`
The total charge of the battery is given by
`Q = C_"net"V = (4 "uF") xx (100 "V") = 4 xx 10^-4 C`
For a and d,
`q = 4 xx 10^-4 "C" and "C" = 10^-5 "F"`
`therefore E = q^2/(2C) = (4 xx 10^-4)/(2 xx 10^-5)`
`⇒ E = 8 xx 10^-3 "J" = 8 "mJ"`
For b and c,
`q = 4 xx 10^-4 "C" and "C"_(eq) = 2"C" = 2 xx 10^-5 "F"`
`therefore V = q/C_(eq) (4 xx 10^-4)/(2 xx 10^-5) = 20 V`
`⇒ E = 1/2 CV^2`
`⇒ E = 1/2 xx 10^-5 xx 400`
`⇒ E = 2 xx 10^-3 "J" = 2 "mJ"`
