AP EAMCET201726 Apr 2017Morning ShiftPhysicsCurrent ElectricityActual
In steady state, a capacitor of capacitance 2 F is charged to 4 C , as shown in figure. If the internal resistance of the cell is 0.5 , then the emf of the cell is
Options
- A4 ~V
- B5 ~V
- C2.5 ~V
- D2 ~V
Correct answer
C. 2.5 ~V
Step-by-step solution
V _C= Q C = 4 10⁻⁶ 2 10⁻⁶ =2 v Now, V _C= V (across 2 resistor) I= V R = 2 2 =1 ~A Now, using the relation, V =E-I r So, E =V+I r=2+1 0.5=2.5 ~V