Question

Three capacitors having capacitances of 8.4, 8.4, and 4.2 \(\mu\)F are connected in series across a 36-V potential difference. (a) What is the charge on the 4.2-\(\mu\)F capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

Short answer

(a) 75.6 µC, (b) 1.36 mJ, (c) 3.6 V, (d) 0.136 mJ.

Step-by-step solution

Understanding Capacitors in Series

In a series connection, the charge \( Q \) on each capacitor is the same, and the total voltage \( V_{total} \) is the sum of voltages across each capacitor. The equivalent capacitance \( C_{eq} \) for capacitors in series is given by: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]For our capacitors: \[ \frac{1}{C_{eq}} = \frac{1}{8.4} + \frac{1}{8.4} + \frac{1}{4.2} \]"Calculating” and solving, we find \( C_{eq} = 2.1 \, \mu\text{F} \)

Calculating Charge on the 4.2 µF Capacitor

For capacitors in series, the charge \( Q \) is same and calculated from \( Q = C_{eq} \times V_{total} \). Substituting the values, we get:\[ Q = 2.1 \, \mu\text{F} \times 36 \, \text{V} = 75.6 \, \mu\text{C} \]Thus, the charge on the 4.2 µF capacitor is 75.6 µC.

Calculating Total Energy Stored in Capacitors in Series

The energy \( E \) stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \). Using the equivalent capacitance and voltage:\[ E_{total} = \frac{1}{2} \times 2.1 \, \mu\text{F} \times (36 \, \text{V})^2 = 1.36 \, \text{mJ} \]This is the total energy stored when capacitors are in series.

Reconfiguring Capacitors in Parallel

When the capacitors are reconnected in parallel, each has the same charge as before: 75.6 µC. The total capacitance \( C_{total} \) in parallel is:\[ C_{total} = C_1 + C_2 + C_3 = 8.4 + 8.4 + 4.2 = 21 \, \mu\text{F} \]Voltage across each capacitor in parallel is given by: \[ V = \frac{Q_{total}}{C_{total}} = \frac{75.6 \, \mu\text{C}}{21 \, \mu\text{F}} = 3.6 \, \text{V} \]

Calculating Total Energy Stored after Reconfiguration

The new total energy stored in the parallel configuration is:\[ E = \frac{1}{2} C_{total} V^2 = \frac{1}{2} \times 21 \, \mu\text{F} \times (3.6 \, \text{V})^2 = 0.136 \, \text{mJ} \]Note that energy is conserved but redistributed among capacitors during reconfiguration.

Key concepts

Capacitance

Capacitance is the ability of a system to store an electric charge. It is a crucial property of capacitors, which are devices that can store energy in the electric field between their plates. The capacitance of a capacitor is defined by the formula: \[ C = \frac{Q}{V} \]where:

• \( C \) is the capacitance
• \( Q \) is the charge stored
• \( V \) is the voltage across the capacitor

In essence, capacitance tells us how much charge a capacitor can hold per unit of voltage applied. It is measured in farads (F), although practical capacitors often use smaller units like microfarads (\( \mu F \)), nanofarads (nF), or picofarads (pF). Having a higher capacitance means that the capacitor can store more charge.
Understanding capacitance is essential when working with circuits, as it dictates how capacitors will interact with each other and influence the overall behavior of the system.

Energy Stored in Capacitors

The energy stored in a capacitor arises from the electric field between its plates when they are charged. This stored energy can be used in various applications, such as in flash photography or in power supplies for electronic devices. The formula for calculating the energy \( E \) stored in a capacitor is:\[ E = \frac{1}{2} C V^2 \]where:

• \( E \) is the energy in joules (J)
• \( C \) is the capacitance in farads
• \( V \) is the voltage across the capacitor in volts

In the given exercise, by connecting capacitors in different configurations, we observe changes in the total energy stored. Initially, when capacitors are in series, their energy is calculated using the combined equivalent capacitance and the total applied voltage. By rearranging them in parallel, while maintaining the same charge, the energy stored changes due to the variation in voltage across each capacitor.
This shift in energy is crucial for understanding how capacitors can be utilized efficiently in various circuit designs.

Capacitors in Parallel

When capacitors are connected in parallel, they are arranged such that each capacitor's positive terminal is connected to the positive terminal of the power source, and the negative terminals are also connected together. This configuration ensures that the voltage across each capacitor is identical, but their charges can differ. The total capacitance \( C_{total} \) of capacitors in parallel is a straightforward sum:\[ C_{total} = C_1 + C_2 + C_3 + ... \]This means the total capacitance is the sum of the individual capacitances.
In the exercise scenario, when the capacitors were reconfigured to be in parallel with the same initial charge, the new total capacitance greatly influenced the voltage across each capacitor. By summing the capacitance of individual capacitors, their collective ability to store charge increases.
This configuration is particularly useful in circuits where a higher total capacitance is beneficial, permitting a larger reserve of stored energy at the same applied voltage. Understanding how capacitors add up in parallel is essential for designing circuits that can handle variable energy demands.