Cylindrical and Spherical Capacitors: Series and Parallel Combinations
- Capacitors can be connected in series and parallel combinations
- In series combination, the total capacitance is less than the capacitance of individual capacitors
- In parallel combination, the total capacitance is the sum of the capacitance of individual capacitors
- Let’s derive the formula for the series combination
Consider two capacitors connected in series with capacitances C1 and C2.
- The charge on the capacitors is the same, denoted by Q.
- The voltage across C1 is V1 and across C2 is V2.
- Applying the fundamental equation Q = CV, we have Q = C1 * V1 and Q = C2 * V2.
- The total voltage V across the series combination is the sum of individual voltages: V = V1 + V2.
- From the fundamental equation, we have Q = CT * V, where CT is the total capacitance.
- Substituting the values of Q, V, C1, and C2, we obtain C1 * V1 + C2 * V2 = CT * (V1 + V2).
- Rearranging the equation gives us C1 * V1 + C2 * V2 = CT * V1 + CT * V2.
- Grouping terms, we have C1 * V1 - CT * V1 = CT * V2 - C2 * V2.
- Factoring out V1 and V2, we obtain V1 * (C1 - CT) = V2 * (CT - C2).
- Dividing both sides by CT - C2, we get V1 = (V2 * (CT - C2)) / (C1 - CT).
Series Combination: Example
Consider two capacitors with capacitances C1 = 4 µF and C2 = 6 µF.
- The total capacitance in the series combination is CT = (C1 * C2) / (C1 + C2).
- Substituting the given values, CT = (4 µF * 6 µF) / (4 µF + 6 µF) = 24 µF / 10 µF = 2.4 µF.
- Let’s assume V2 = 10V.
- We can now calculate V1 using the formula derived previously.
Series Combination: Example (2)
- Substituting V2 = 10V, CT = 2.4 µF, C1 = 4 µF, and C2 = 6 µF in the formula, we have V1 = (10V * (2.4 µF - 6 µF)) / (4 µF - 2.4 µF).
- Simplifying the equation, V1 = (10V * (-3.6 µF)) / (1.6 µF) = -22.5V.
- Therefore, the voltage across the first capacitor (V1) is -22.5V, indicating that its polarity is opposite to the chosen direction.
- Capacitors can also be connected in parallel.
- In this case, the total capacitance (CT) is the sum of individual capacitances: CT = C1 + C2 + C3 + …
- The voltage across each capacitor is the same in a parallel combination.
Parallel Combination: Example
Consider three capacitors with capacitances C1 = 2 µF, C2 = 3 µF, and C3 = 4 µF.
- The total capacitance in the parallel combination is CT = C1 + C2 + C3.
- Substituting the given values, CT = 2 µF + 3 µF + 4 µF = 9 µF.
- The voltage across each capacitor in a parallel combination is the same.
Parallel Combination: Example (2)
- Let’s assume V = 8V, which is the voltage across each capacitor in the parallel combination.
- The charge on each capacitor can be calculated using the fundamental equation Q = CV.
- Q1 = C1 * V = 2 µF * 8V = 16 μC
- Q2 = C2 * V = 3 µF * 8V = 24 μC
- Q3 = C3 * V = 4 µF * 8V = 32 μC
Summary
- Capacitors can be connected in series and parallel combinations.
- In series combination, the total capacitance is less than the capacitance of individual capacitors.
- In parallel combination, the total capacitance is the sum of the capacitance of individual capacitors.
- The formula for the series combination is derived using the fundamental equation Q = CV.
- Examples and equations were provided to illustrate the concepts.
Series Combination: Recap
- Capacitors connected in series have a total capacitance less than the individual capacitors.
- The voltage across each capacitor in a series combination is different.
- The formulas we derived for the series combination are:
- CT = (C1 * C2) / (C1 + C2)
- V1 = (V2 * (CT - C2)) / (C1 - CT)
Parallel Combination: Recap
- Capacitors connected in parallel have a total capacitance equal to the sum of individual capacitances.
- The voltage across each capacitor in a parallel combination is the same.
- Let’s consider an example to understand parallel combination better.
Parallel Combination: Example
Consider two capacitors with capacitances C1 = 2 µF and C2 = 4 µF.
- The total capacitance in the parallel combination is CT = C1 + C2.
- Substituting the given values, CT = 2 µF + 4 µF = 6 µF.
- Let’s assume the voltage across each capacitor (V) is 12V.
- Using the fundamental equation Q = CV, we can calculate the charge on each capacitor.
Parallel Combination: Example (2)
- The charge on each capacitor can be calculated using the formula Q = CV.
- Q1 = C1 * V = 2 µF * 12V = 24 μC
- Q2 = C2 * V = 4 µF * 12V = 48 μC
- The total charge in the parallel combination is the sum of individual charges.
Parallel Combination: Example (3)
- The total charge in the parallel combination is Q = Q1 + Q2.
- Substituting the calculated values, Q = 24 μC + 48 μC = 72 μC.
- The voltage across each capacitor in a parallel combination is the same, whereas the charge on each capacitor varies.
Series and Parallel Combination: Capacitance Ratio
- When capacitors are connected in series, the total capacitance is smaller than the individual capacitors.
- For capacitors connected in parallel, the total capacitance is greater than any individual capacitance.
- The ratio of total capacitance in series to total capacitance in parallel is given by the expression:
CT(series) / CT(parallel) = C1 + C2 + C3 + … / (C1 * C2 * C3 * …)
- This ratio illustrates the relationship between series and parallel combinations.
Applications of Series Combinations
- Series combinations of capacitors are used in various applications, such as:
- Power factor correction circuits
- Filtering circuits in audio systems
- Electronic timers and oscillators
- Understanding series combinations helps in designing and analyzing these circuits properly.
Applications of Parallel Combinations
- Parallel combinations of capacitors find applications in several areas, including:
- Energy storage devices in power supply circuits
- Smoothing circuits in power converters
- High-frequency decoupling capacitors in electronic systems
- Knowing the behavior of parallel combinations aids in optimizing circuit performance.
Recap and Summary
- Capacitors can be connected in series and parallel combinations.
- In series combinations, the total capacitance is smaller, and the voltage across each capacitor varies.
- In parallel combinations, the total capacitance is larger, and the voltage across each capacitor is the same.
- Ratios of capacitance in series and parallel combinations are also important in understanding circuits.
- Applications of series and parallel combinations are found in various electronic systems.
Questions?
Do you have any questions related to the topics covered today? Feel free to ask for clarification or further explanation.
Cylindrical and Spherical Capacitors
- Capacitors come in different shapes and sizes.
- Cylindrical capacitors have a cylindrical shape with conductive plates on the outer and inner surfaces.
- Spherical capacitors have concentric spherical conductive shells.
Consider two cylindrical capacitors connected in series.
- The capacitance of a cylindrical capacitor can be calculated using the formula C = 2πε₀l / ln(b/a).
- Let’s assume the two capacitors have lengths l1 and l2, and radii a1, a2, b1, and b2 for the outer and inner cylinders respectively.
- The total capacitance in series combination is CT = (C1 * C2) / (C1 + C2).
- Substituting the formulas for the capacitance, we have CT = [(2πε₀l1 / ln(b1/a1)) * (2πε₀l2 / ln(b2/a2))] / [(2πε₀l1 / ln(b1/a1)) + (2πε₀l2 / ln(b2/a2))].
- Simplifying the equation gives us CT = [(l1 * l2) / ((ln(b1/a1) * ln(b2/a2)) / (ln(b1/a1) + ln(b2/a2)))].
Series Combination: Example
Consider two cylindrical capacitors with lengths l1 = 10 cm, l2 = 15 cm, radii a1 = 2 cm, a2 = 3 cm, b1 = 5 cm, and b2 = 6 cm.
- Using the derived formula CT = [(l1 * l2) / ((ln(b1/a1) * ln(b2/a2)) / (ln(b1/a1) + ln(b2/a2)))], we can calculate the total capacitance in the series combination.
Series Combination: Example (2)
- Substituting the given values, CT = [(10 cm * 15 cm) / ((ln(5 cm/2 cm) * ln(6 cm/3 cm)) / (ln(5 cm/2 cm) + ln(6 cm/3 cm)))].
- Simplifying the equation, CT = [(150 cm²) / ((1.609 * 1.099) / (1.609 + 1.099))] ≈ 140.9 cm².
- Therefore, the total capacitance in the series combination is approximately 140.9 cm².
Consider two cylindrical capacitors connected in parallel.
- The total capacitance in the parallel combination is the sum of individual capacitances: CT = C1 + C2.
- Substituting the formulas for the capacitance, we have CT = (2πε₀l1 / ln(b1/a1)) + (2πε₀l2 / ln(b2/a2)).
Parallel Combination: Example
Continuing with the example of two cylindrical capacitors:
- Capacitor 1 has a length l1 = 10 cm, radius a1 = 2 cm, and radius b1 = 5 cm.
- Capacitor 2 has a length l2 = 15 cm, radius a2 = 3 cm, and radius b2 = 6 cm.
- Using the formula CT = (2πε₀l1 / ln(b1/a1)) + (2πε₀l2 / ln(b2/a2)), we can calculate the total capacitance in the parallel combination.
Parallel Combination: Example (2)
- Substituting the given values, CT = (2πε₀ * 10 cm / ln(5 cm/2 cm)) + (2πε₀ * 15 cm / ln(6 cm/3 cm)).
- Simplifying the equation, CT = (20πε₀ / ln(2.5)) + (30πε₀ / ln(2)).
- Therefore, the total capacitance in the parallel combination is (20πε₀ / ln(2.5)) + (30πε₀ / ln(2)).
Recap and Summary
- Cylindrical capacitors have a cylindrical shape with conductive plates on the outer and inner surfaces.
- The formulas for the capacitance of cylindrical capacitors are C = 2πε₀l / ln(b/a).
- Series combination formula for cylindrical capacitors is CT = [(l1 * l2) / ((ln(b1/a1) * ln(b2/a2)) / (ln(b1/a1) + ln(b2/a2)))].
- Parallel combination formula for cylindrical capacitors is CT = (2πε₀l1 / ln(b1/a1)) + (2πε₀l2 / ln(b2/a2)).
- Examples were provided to illustrate the usage of the formulas.
Questions?
Do you have any questions related to the topics covered today? Feel free to ask for clarification or further explanation.