A capacitor of capacitance 5 F is connected as shown in the figure. The internal resistance of the cell is 0.5 . The amount of charge on the capacitor plates is

Step-by-step solution

In steady state, there will be no current in the capacitor branch. Net resistance of the circuit R=1+1+0.5=2.5 Current drawn from the cell, i= V R = 2.5 2.5 =1 ~A Potential drop across two parallel branches aligned V=E-i r & =2.5-1 0.5 & =2.5-0.5 & =2.0 ~V aligned So, charge on the capacitor plates aligned q=C V & =5 2 & =10 C aligned