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Question
For the given capacitor configuration
- Find the charges on each capacitor
- potential difference across them
- energy stored in each capacitor.
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Solution
Capacitor b and c in parallel combination
Cp = Cb + Cc = (6 + 2) μF = 8 μF
Capacitor a, cp 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
- Charge on each capacitor,
Charge on capacitor a, Qa = Cs V = `8/3 xx 9`
Qa = 24 μC
Charge on capacitor, d, Qd = Cs V = `8/3 xx 9`
Qd = 24 μC
Capacitor b and c in parallel
Charge on capacitor, b, Qb = `6/3 xx 9` = 18
Qb = 18 μC
Charge on capacitor, c, Qc = `2/3 xx 9` = 6
Qc = 6 μC - Potential difference across each capacitor, V = `"q"/"c"`
Capacitor Ca, Va = `"q"_"a"/"C"_"a" = (24 xx 10^6)/(8 xx 10^6)` = 3 V
Capacitor Cb, Vb = `"q"_"b"/"C"_"b" = (18 xx 10^6)/(6 xx 10^6)` = 3 V
Capacitor Cc, Vc = `"q"_"c"/"C"_"c" = (6 xx 10^6)/(2 xx 10^6)` = 3 V
Capacitor Cd, Vd = `"q"_"d"/"C"_"d" = (24 xx 10^6)/(8 xx 10^6)` = 3 V - Energy stores in a capacitor, U = `1/2` CV2
Energy in capacitor Ca, Ua = `1/2 "C"_"a" "V"_"a"^2 = 1/2 xx 8 xx 10^-6 xx (3)^2`
Ua = 36 μJ
Capacitor Cb, Ub = `1/2 "C"_"b" "V"_"b"^2 = 1/2 xx 6 xx 10^-6 xx (3)^2`
Ua = 27 μJ
Cc, Uc = `1/2 "C"_"c" "V"_"c"^2 = 1/2 xx 2 xx 10^-6 xx (3)^2`
Ua = 9 μJ
Cd, Ud = `1/2 "C"_"d" "V"_"d"^2 = 1/2 xx 8 xx 10^-6 xx (3)^2`
Ua = 36 μJ
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