I have the correct answers already the most important part is the work and expla

Question

I have the correct answers already the most important part is the work and explanation! Please explain or that does me no good!!

in text: In Fig. 24.26 C1= 3.00 uF and Vab = 120 V. the charge pm capacitor C1 is 150 uC. Calculate the voltage across the other two capacitors.

Consider the arrangement of capacitors shown in Fig. 26.26. Find the energy stored in (a) the 5 uF capacitor; (b) the 4-uF capacitor.

I have the correct answers already the most important part is the work and explanation! Please explain or that does me no good!! in text: In Fig. 24.26 C1= 3.00 MuF and Vab = 120 V. the charge pm capacitor C1 is 150 MuC. Calculate the voltage across the other two capacitors. Consider the arrangement of capacitors shown in Fig. 26.26. Find the energy stored in (a) the 5 MuF capacitor; (b) the 4-uF capacitor.

Explanation / Answer

24.26

Charge on C1 = 150 muC

capacitance of C1 = 3muC

So voltage drop across C1 = Q/C1 = 150/3 = 50 V

Now C1and C2 are in parallel, So voltage drop across C2 = 50 V (ans)

VOltage across C3 = 120 - 50 = 70 V (ans)

2nd one)

Net capacitance of 2muF and 4muF = 2*4/6 = 1.333 muF

This capacitance is in paralllel with 12muF = 12+ 1.333 = 13.333 muF

Net capacitance of the circuti = 13.3333*5/18.333 = 3.6364 muF

Net charge in the ciircuit = CV = 20*3.6364 = 72.728 muC

Energy stored by 5 mu F can be given by = Q2/2C    [ 5muF stores the full charge 72.728 muC , charge on capacitors in series is same]                                      

So, Energy = (72.728*10-6)2/2*5*10-6 = 528.9 * 10-6 joules (ans)

Now ,

charge on the upper branch of the circuit = 1.3333* 72.728/ 13.333 = 7.27 muC

Energy stored by 4 muF = (7.27*10-6)2/2*4 = 6.608 *10-6Joules (ans)

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