Question

For the given capacitor configuration

1. Find the charges on each capacitor
2. potential difference across them
3. energy stored in each capacitor.

Answer 1

Capacitor b and c in parallel combination

C_p = C_b + C_c = (6 + 2) μF = 8 μF

Capacitor a, c_p and d are in series combination, so the resulatant copacitance

`1/"C"_"s" = 1/"C"_"a" + 1/"C"_"cp" + 1/"C"_"d"`

`= 1/8 + 1/8 + 1/8 = 3/8`

`"C"_"s" = 8/3` μF

1. Charge on each capacitor,
   Charge on capacitor a, Q_a = C_s V = `8/3 xx 9`
   Q_a = 24 μC
   Charge on capacitor, d, Q_d = C_s V = `8/3 xx 9`
   Q_d = 24 μC
   Capacitor b and c in parallel
   Charge on capacitor, b, Q_b = `6/3 xx 9` = 18
   Q_b = 18 μC
   Charge on capacitor, c, Q_c = `2/3 xx 9` = 6
   Q_c = 6 μC
2. Potential difference across each capacitor, V = `"q"/"c"`
   Capacitor C_a, V_a = `"q"_"a"/"C"_"a" = (24 xx 10^6)/(8 xx 10^6)` = 3 V
   Capacitor C_b, V_b = `"q"_"b"/"C"_"b" = (18 xx 10^6)/(6 xx 10^6)` = 3 V
   Capacitor C_c, V_c = `"q"_"c"/"C"_"c" = (6 xx 10^6)/(2 xx 10^6)` = 3 V
   Capacitor C_d, V_d = `"q"_"d"/"C"_"d" = (24 xx 10^6)/(8 xx 10^6)` = 3 V
3. Energy stores in a capacitor, U = `1/2` CV²
   Energy in capacitor C_a, U_a = `1/2 "C"_"a" "V"_"a"^2 = 1/2 xx 8 xx 10^-6 xx (3)^2`
   U_a = 36 μJ
   Capacitor C_b, U_b = `1/2 "C"_"b"  "V"_"b"^2 = 1/2 xx 6 xx 10^-6 xx (3)^2`
   U_a = 27 μJ
   C_c, U_c = `1/2 "C"_"c"  "V"_"c"^2 = 1/2 xx 2 xx 10^-6 xx (3)^2`
   U_a = 9 μJ
   C_d, U_d = `1/2 "C"_"d"  "V"_"d"^2 = 1/2 xx 8 xx 10^-6 xx (3)^2`
   

   U_a = 36 μJ