99 Percentile Qs Bank for JEE MainPhysicsCapacitance
Four capacitors marked with capacitances and breakdown voltages are connected as shown in the figure. The maximum emf of the source, so that no capacitor breaks down is
Options
- A10.5 kV
- B5.25 kV
- C2.25 kV
- D1.25 kV
Correct answer
C. 2.25 kV
Step-by-step solution
Resultant capacitance of series combination 1 , aligned & = 1 5 + 1 4 = 9 20 C_ eq ₁ & = 20 9 =2.25 F aligned Resultant capacitance of series combination 2, aligned & = 1 3 + 1 2 = 5 6 C_ eq ₂ & = 6 5 =1 2 F aligned So, charge on upper branch is = 20 9 ~V and charge on lower branch is 6 5 ~V . So, smallest value is 2.25 kV.