Saved Bookmarks
| 1. |
A network of four capacitors, each of capacitance 15muF,is connected across a battery of 100 V as shown in the figure. Find the net capacitance and the charge on the capacitor C_4. |
|
Answer» Solution :Here `C_1 = C_2 = C_3 = C_4= 15 muF and V = 100V` In the network shown,capacitors `C_1, C_2 andC_3` are joined in series and their equivalent capacitance C. is GIVEN by : `1/(C.) = 1/C_1 + 1/C_2 + 1/C_3 = 1/15+ 1/15+ 1/15 = 1/5 rArr C. = 5 muF`. Now, C. and `C_4` are connected in parallel, hence the net capacitance of the network : `C = C. + C_4= (5 + 15) mu F= 20 muF` Moreover, the charge on the capacitor `C_4`, `Q_4= C_4 V = (15 muF) xx 100V =1500 muC = 1.5 MC` |
|