Question

Three capacitors are connected to a 12 volt battery as shown above. Given the values of the capacitors, find the requested information below. C = 1.72uF C = 4.86F C3 = 2.61uF (A) What is the total capacitance of the circuit in uF? (B) How much energy is stored in the circuit? (C) What is the voltage across each capacitor? 3 III

Answer 1

DELTA/Page No. Date 1 C, = 1.72 uf C = 4.86 4f G = 2.61 uf H tal Total Capactance Ca and Co are in series, so their equuilent Capacitor is = CS ☆ 1-1 s 4.86 + 1 2.61 = 2161 +4.86 4.86X2161 = 4.86 x 2.67 2.6144.86 & S = 1.698 a MF parallel, so and now cs and C, are in Total capactance CT = (sta C = 1.698 +1.72 CT = 3.418 MF Total capacitance is 3.418 un

DELTA/Page No. w Energy stored in the ciraut o E-LQV = ICCUIV=CV? ; =CV E = 1 car2 E= 4x(3.418) X(12) - Ez246.1 T Energy of the circuit is 246.1 T CC) Voltage accross each capactor- all these capactor in parallel in circult AB So capacitance in the capactor c, I will have same value of voltage as the voltage of circut because in parallel, ciri combination voltage remain same So TV = 12V 2 V 1 CatC37 4.86 (4.86+2.61 TV2 = 7.80 V C3 Cat C 12.61 (2.61+4.86 X12 - 4.19 v

DELTA Papua Date V=4.195 (0) charge on a =) 0 = C,V, - 1:72x12x10 0 = 20.64 MC charge on a P = Egov - 16984427 De = 1:698712 192=20.376 MC same for Cz =) (s. V = 1.698 X 12 =20. 376 MC Az=20.376 uc So, we can see that in a series combination charge semains some on both cabochon 9 = 20.64 uc 9 = 20.376 de & -20.376 uc
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