Consider the following, (Let C_1 = 9.20 mu F and c_2 = 3.20 mu F.) (a) Find the
ID: 1637328 • Letter: C
Question
Consider the following, (Let C_1 = 9.20 mu F and c_2 = 3.20 mu F.) (a) Find the equivalent capacitance of the capacitors in the figure. 306 mu F (b) Find the charge on each capacitor. on the right 9.20 mu F capacitor 28.1 mu C on the left 9.20 mu F capacitor 28.1 mu C on the 3.20 mu F capacitor 9.8 Your response is within 10% of the correct value, This may be due to roundoff error, or you intermediate result to at least four-digit accuracy to minimize roundoff error. mu C on the 6.00 mu F capacitor 18.3 mu C (c) Find the potential difference across each capacitor: on the right 9.20 mu F capacitor 3.05 V on the right 9.20 mu F capacitor 3.05 V on the right 3.20 mu F capacitor 3.05 VExplanation / Answer
As combination of 6.00 F and 3.20 F result in equivalent of 9.20 F and it is in series with other capacitors so charge on the equivalent is 28.1 C. (as per answer of the parts before though as per my calculation the value of charge on combination is Q = 9.0x3.06x10-6 C = 27.54 C)
Then as 6.00 F and 3.20 F are in parallel and voltage is same across them so charge on them is proportional to their capacities and using charge division, charge on 3.20 F is
Q’ = {3.20/(3.20+6.00)}*28.1 = 9.774 C (as per answer of the parts before)
Q’ = {3.20/(3.20+6.00)}*27.54= 9.579 C (as per my calculation)
(Note: as per my calculation the value of charge on combination is Q = 9.0x3.06x10-6 C = 27.54 C and hence charge on 3.20 F is 9.579 C so please try this value)