wh U2-Qua . Chap 24 ?e For The Circuit Shown In myLTL-Blackboard Lee x / -? www.
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wh U2-Qua . Chap 24 ?e For The Circuit Shown In myLTL-Blackboard Lee x / -? www.webassi × eBook 3. -12 points Tipler8 21.P051 My Notes Ask Your Two capacitors, one hat has a capacitance o 8 ??and one that has a capacitance o 24 ?F are first discharged and then are connected in series. The series combination is then connected across he terminals o a 14-V battery. Next, they are carefully disconnectad so that they are not discharged and they are then reconnected to each other positive plate to posltive plate and negative plate to negative plate (a) Firid the potential dillererice across each capscitor after they are retoniettet. V (8 ?F capacitor) V (74 ?F capacitor) (b) Find Uhe energy sedlhe apailor befre Lhey e disonnected rhe tey, ad i Lhe eneruy sled al hey are etunnecle (before disconnected) eBook 4. -12 points Tipless 24 P056 My Notes Ask Your A 14-HF pacitor and a copacitor of unkowrn capecitance are both charged to 2.75 kV. After charuing, the two capatilors are disconnected om the vellage source. The opacs are ther connetted to each other positive plate to negative plate and negative plate to positive plate. The final voltage across the terminals of the 14 ?F capacitor is 1.00 kV. (a) Whal is lhe capil ol the sexund capacilur7 (b) How much enery is dissipated when the conection was made? eBook 3:36 PM 4/6/2018Explanation / Answer
3)
Given,
C1 = 8 uF ; C2 = 24 uF ; V = 14 V
when in series, the equivalent capacitance is:
Ceq = 8 x 24/(8 + 24) = 6 uF
Q = CV = 6 x 10^-6 x 14 = 84 uC
when in parallel
Ceq = 24 + 8 = 32 uF
Q = CV = 2 x 84 = 168 uC
a)V(8) = 168/32 = 5.25 V
V(24) = 168/32 = 5.25 V
b)U = 1/2 C V^2
Ui = 0.5 x 6 x 10^-6 x 14^2 = 5.88 x 10^-4 J
Uf = 0.5 x 32 x 10^-6 x 5.25^2 = 4.41 x 10^-4 J
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