Concepts Of Physics MCQ Edition [Volume 2]PhysicsCapacitors
Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Determine the potential difference across the 20.0 pF and 50.0 pF capacitors respectively.
Options
- A6.00 V and 6.00 V
- B3.00 V and 3.00 V
- C1.71 V and 4.29 V
- D4.29 V and 1.71 V
Correct answer
D. 4.29 V and 1.71 V
Step-by-step solution
Given C₁ = 20.0 pF and C₂ = 50.0 pF connected in series with V = 6.00 V . In a series combination, the charge Q on each capacitor is the same. The equivalent capacitance is C_ eq = C₁ C₂ C₁ + C₂ = 20.0 50.0 20.0 + 50.0 = 100 7 pF . Total charge Q = C_ eq V = 100 7 6.00 = 600 7 pC . Potential difference across the 20.0 pF capacitor is V₁ = Q C₁ = 600/7 20.0 = 30 7 4.29 V . Potential difference across the 50.0 pF capacitor is V₂ = Q C₂ = 600/7 50.0 = 12 7 1.71 V . Alternatively, using the voltage divider rule: V₁ = V