Answer:
1. Voltage across 600 pF is 10 V.
2. Voltage across 300 pF is 20 V.
3. Voltage across 200 pF is 30 V.
Explanation:
We'll begin by calculating the total capacitance of capacitor. This can be obtained as follow:
Capicitance 1 (C₁) = 600 pF
Capicitance 2 (C₂) = 300 pF
Capicitance 3 (C₃) = 200 pF
Total capacitance (Cₜ) =?
1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃
1/Cₜ = 1/600 + 1/300 + 1/200
1/Cₜ = 1 + 2 + 3 / 600
1/Cₜ = 6/600
1/Cₜ = 1/100
Cₜ = 100 pF
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Thus, 100 pF is equivalent to 1×10¯¹⁰ F.
Next, we shall determine the charge. This can be obtained as follow:
Voltage (V) = 60 V
Capicitance (C) = 1×10¯¹⁰ F
Charge (Q) =?
Q = CV
Q = 60 × 1×10¯¹⁰ F
Q = 6×10¯⁹ C
1. Determination of the voltage across 600 pF.
Capicitance 1 (C₁) = 600 pF = 6×10¯¹⁰ F
Charge (Q) = 6×10¯⁹ C
Voltage 1 (V₁) =?
Q = C₁V₁
6×10¯⁹ = 6×10¯¹⁰ × V₁
Divide both side by 6×10¯¹⁰
V₁ = 6×10¯⁹ / 6×10¯¹⁰
V₁ = 10 V
2. Determination of the voltage across 300 pF.
Capicitance 2 (C₂) = 300 pF = 3×10¯¹⁰ F
Charge (Q) = 6×10¯⁹ C
Voltage 2 (V₂) =?
Q = C₂V₂
6×10¯⁹ = 3×10¯¹⁰ × V₂
Divide both side by 3×10¯¹⁰
V₂ = 6×10¯⁹ / 3×10¯¹⁰
V₂ = 20 V
3. Determination of the voltage across 200 pF.
Capicitance 3 (C₃) = 200 pF = 2×10¯¹⁰ F
Charge (Q) = 6×10¯⁹ C
Voltage 3 (V₃) =?
Q = C₃V₃
6×10¯⁹ = 2×10¯¹⁰ × V₃
Divide both side by 2×10¯¹⁰
V₃ = 6×10¯⁹ / 2×10¯¹⁰
V₃ = 30 V
