Problem 61 - University Physics with Modern Physics

Question

Three capacitors having capacitances of \(8.4,8.4,\) and 4.2\(\mu \mathrm{F}\) are connected in series across a \(36-\mathrm{V}\) 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?

Step-by-Step Solution

Answer

(a) 100.8 μC; (b) 1.814 mJ; (c) 14.4 V; (d) 2.088 mJ.

Step 1: Find Equivalent Capacitance

For capacitors in series, the equivalent capacitance \(C_{eq}\) is given by: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] where \(C_1 = 8.4\, \mu F\), \(C_2 = 8.4\, \mu F\), and \(C_3 = 4.2\, \mu F\). Calculating this, we get: \[ \frac{1}{C_{eq}} = \frac{1}{8.4} + \frac{1}{8.4} + \frac{1}{4.2} = 0.3571\, \mu F^{-1} \] Thus, \(C_{eq} \approx 2.8\, \mu F\).

Step 2: Calculate Charge on Each Capacitor

The charge \(Q\) on capacitors in series is the same. Using \(Q = C_{eq} \cdot V\), where \(V = 36\, V\): \[ Q = 2.8\, \mu F \times 36\, V = 100.8\, \mu C \] Thus, the charge on the \(4.2\, \mu F\) capacitor is also \(100.8\, \mu C\).

Step 3: Calculate Total Energy in Series Configuration

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\). For the series, the energy is: \[ E_{total} = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times 2.8\, \mu F \times (36\, V)^2 = 1.814\, mJ \]

Step 4: Find Voltage Across Each Capacitor in Parallel

When disconnected and reconnected in parallel without discharging, the voltage across each capacitor is found using total charge \(Q_{total} = Q \times 3\) and total capacitance. \(Q_{total} = 100.8\, \mu C \times 3 = 302.4\, \mu C\), and \(C_{total} = C_1 + C_2 + C_3 = 21\, \mu F\). Thus, voltage \(V'\) is: \[ V' = \frac{Q_{total}}{C_{total}} = \frac{302.4\, \mu C}{21\, \mu F} \approx 14.4\, V \]

Step 5: Calculate Total Energy in Parallel Configuration

In the parallel combination, the energy stored in each capacitor is \(E = \frac{1}{2} C V'^2\). Total energy \(E'\) is: \[ E' = 3 \times \frac{1}{2} C_1 (V')^2 = 3 \times \frac{1}{2} \times 8.4\, \mu F \times (14.4 \ V)^2 = 2.088\, mJ \]

Key Concepts

Equivalent Capacitance

When capacitors are connected in series, the equivalent capacitance of the system can be found using the reciprocal formula: \[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\]This formula shows that the total capacitance for capacitors in series is always less than the smallest capacitance in the series.

• It sums the reciprocals of the individual capacitances.
• The calculation ensures that capacitance diminishes because, in a series, the same charge must pass through each capacitor.

For instance, if you have capacitances of 8.4, 8.4, and 4.2 microfarads, the equivalent capacitance would be approximately 2.8 microfarads. Calculating the equivalent capacitance in series is crucial because it allows you to explore how capacitors share charge and energy across a circuit. Understanding this lays a foundation for working with complex circuits involving multiple capacitors.

Charge on Capacitors

In a series circuit, the charge \(Q\) across each capacitor is the same. This is because the current flowing through the circuit must pass through each capacitor, rendering each charge carry equal amounts of electron flow. The total charge in this case is calculated by:\[Q = C_{eq} \cdot V\]where \(C_{eq}\) is the equivalent capacitance, and \(V\) is the applied potential difference across the series. For a 36 V potential difference, using our calculated \(C_{eq} = 2.8 \mu F\), the charge \(Q\) on each capacitor is approximately 100.8 μC.

• Always remember, in series circuits, though the charge is identical on all capacitors, the potential difference (voltage) can vary.
• This characteristic is fundamental when designing circuits that require specific charge distributions.

It is critical to grasp this concept as it influences how capacitors behave in different circuit configurations.

Energy Stored in Capacitors

Capacitors store energy in the electric field between their plates. The energy \(E\) stored in a capacitor, when connected in series, can be determined using:\[E = \frac{1}{2} C_{eq} V^2\]where \(V\) is the voltage. For the total series configuration, with a voltage of 36 V, the energy stored would be approximately 1.814 millijoules.

• Energy storage is a vital feature in applications like power supply systems where capacitors smooth out voltage fluctuations.
• Understanding energy calculation helps in ensuring the safe and efficient design of circuits.

Analyzing stored energy assists in efficiently managing power within electronic systems.

Voltage Across Capacitors in Parallel

When capacitors that were originally in series are reconfigured to be in parallel, without discharging, they will share a common voltage \(V'\) across them. This is because the capacitors are essentially connected to the same two points in a circuit, ensuring identical electrical potential across each. Calculate the new voltage across each capacitor in the parallel setup by using:\[V' = \frac{Q_{total}}{C_{total}}\]Here, \(Q_{total}\) is the pre-calculated total charge, and \(C_{total}\) is the sum of the individual capacitances now in parallel. For example, with a total charge of 302.4 µC and a total capacitance of 21 µF, the voltage \(V'\) comes out to be 14.4 V.

• This uniform voltage facilitates equal energy distribution across each capacitor.
• In parallel, capacitors can supply energy or smooth voltage efficiently, which is beneficial in many electronic applications.

Understanding this characteristic is important for tailoring capacitive circuits for specific needs and functionalities.