Saved Bookmarks
| 1. |
A capacitor of capacitance 5μF is connected as shown in the figure. The internal resistance of the cell is 0.5Ω. The amount of charge on the capacitor plates is? |
|
Answer» In steady state, there will be no current in the capacitor branch. Net resistance of the circuit R = 1+1+0.5 = 2.5Ω Current drawn from the cell i = RV = 2.52.5 = 1A Potential drop across two parallel branches V = E−ir = 2.5−1×0.5 = 2.5−0.5 = 2.0V So, charge on the capacitor plates q = CV = 5×2 = 10μC. |
|