Consider the circuit shown in the figure, with C1=6.52μF and C2=8.54μFμF [Free] B118
Consider the circuit shown in the figure, with C1=6.52μF and C2=8.54μFμF μFμC 6.52μF capacitor μC 6.00μF capacitor μC 8.54μF capacitor μC 2.00μF capacitor μC (c) Find the potential difference (in V ) on each capacitor. 6.52μF capacitor V 6.00μF capacitor V 8.54μF capacitor V 2.00 uF capacitor V
![Consider the circuit shown in the figure, with C1=6.52μF and C2=8.54μFμF [Free] B118](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20640%20642'%3E%3C/svg%3E)
Answer
Capacitor Network Analysis: Series and Parallel Capacitors
This step-by-step tutorial guides you through calculating the equivalent capacitance, charge, and voltage in a capacitor network involving both series and parallel connections.
Step 1: Equivalent Capacitance for Series Capacitors
Capacitors in series follow the rule:
First Series Group:
- C₁ = 6.52µF
- 6µF capacitor
1/Cₓ = 1/6.52 + 1/6
Cₓ = 3.125µF
Second Series Group:
- C₂ = 8.54µF
- 2µF capacitor
1/Cᵧ = 1/8.54 + 1/2
Cᵧ = 1.62µF
Step 2: Parallel Capacitor Combination
For capacitors in parallel:
Ceq = 3.125µF + 1.62µF = 4.745µF
Step 3: Calculate Total Charge
Using the formula:
V = 90V, C = 4.745µF
QT = 4.745 × 10⁻⁶ × 90 = 4.275 × 10⁻⁴ C
Step 4: Charge in Parallel Branches
Since Cₓ and Cᵧ are in parallel, they share the same voltage:
Vₓ = Vᵧ = 90V
Charge on Cₓ:
Charge on Cᵧ:
Step 5: Voltages in Series Capacitors
Group Cₓ: [6.52µF, 6µF]
Charge Q₁ = Q₆ = Qₓ = 2.8125 × 10⁻⁴ C
- V₁ = Q / C = 2.8125e-4 / 6.52e-6 = 43.14V
- V₆ = Q / C = 2.8125e-4 / 6e-6 = 46.87V
Group Cᵧ: [8.54µF, 2µF]
Charge Q₂ = Q₂µF = Qᵧ = 1.458 × 10⁻⁴ C
- V₂ = Q / C = 1.458e-4 / 8.54e-6 = 17.1V
- V₂µF = Q / C = 1.458e-4 / 2e-6 = 72.9V
Final Answers
(a) Equivalent Capacitance:
Ceq = 4.745µF
(b) Charges:
- Q₁ = Q₆ = 2.8125 × 10⁻⁴ C
- Q₂ = Q₂µF = 1.458 × 10⁻⁴ C
(c) Voltages:
- V₁ = 43.14V
- V₆ = 46.87V
- V₂ = 17.1V
- V₂µF = 72.9V
Concepts Used
- Series Capacitors: Reciprocal formula, same charge, divided voltage.
- Parallel Capacitors: Sum of capacitances, same voltage, different charges.
- Q = C × V and V = Q / C used for all calculations.
Use Cases
This analysis is widely applicable in:
- Power supply design
- Signal filtering and conditioning
- Timing and delay circuits
- Energy storage and release mechanisms