Problem 51 - College Physics

Question

(a) Two capacitors can be connected to a battery in either a series or parallel combination. The parallel combination will require (1) more, (2) equal, (3) less energy from a battery than the series combination. Why? (b) Two uncharged capacitors, one with a capacitance of \(0.75 \mu \mathrm{F}\) and the other with that of \(0.30 \mu \mathrm{F}\) are connected in series to a 12-V battery. Then the capacitors are disconnected, discharged, and reconnected to the same battery in parallel. Calculate the energy loss of the battery in both cases.

Step-by-Step Solution

Answer

(a) More energy is required for parallel connection. (b) The energy loss is \(6.06 \times 10^{-6} \, J\).

Step 1: Understand Capacitance in Series and Parallel

When capacitors are connected in series, the total capacitance is given by the formula \( C_s = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} \). In parallel, the total capacitance is the sum: \( C_p = C_1 + C_2 \).

Step 2: Calculate Series Capacitance

Given two capacitors with \( C_1 = 0.75 \, \mu F \) and \( C_2 = 0.30 \, \mu F \), the series capacitance \( C_s \) is calculated as follows:\[ C_s = \frac{1}{\frac{1}{0.75} + \frac{1}{0.30}} \ = 0.214 \, \mu F \]

Step 3: Calculate Parallel Capacitance

For the parallel combination, the capacitance is the sum of the individual capacitances:\[ C_p = 0.75 \, \mu F + 0.30 \, \mu F = 1.05 \, \mu F \]

Step 4: Calculate Energy Stored in Series

The energy stored in capacitors in series is given by \( E_s = \frac{1}{2} C_s V^2 \). Using \( C_s = 0.214 \, \mu F \) and \( V = 12 \, V \):\[ E_s = \frac{1}{2} \times 0.214 \times 10^{-6} \times (12)^2 = 1.5 \times 10^{-6} \, J \]

Step 5: Calculate Energy Stored in Parallel

The energy stored in capacitors in parallel is calculated using \( E_p = \frac{1}{2} C_p V^2 \). With \( C_p = 1.05 \, \mu F \) and \( V = 12 \, V \):\[ E_p = \frac{1}{2} \times 1.05 \times 10^{-6} \times (12)^2 = 7.56 \times 10^{-6} \, J \]

Step 6: Determine Energy Change

The energy loss is the difference in energy stored between the two connections:\[ \Delta E = E_p - E_s = 7.56 \times 10^{-6} \, J - 1.5 \times 10^{-6} \, J = 6.06 \times 10^{-6} \, J \]

Step 7: Conclusion: Comparative Energy Usage

The parallel combination requires more energy from the battery than the series combination, as seen from the calculated energy stored in each configuration.

Key Concepts

Capacitance in Series

Connecting capacitors in series is like creating a longer electrical path for charge to travel. When capacitors are in series, the overall capacitance decreases. To calculate the total capacitance in a series configuration, we use the formula:\[ C_s = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} \]This means the reciprocal of the total capacitance is the sum of the reciprocals of each individual capacitance. It's important to understand that even though the individual capacitances might be large, the series connection results in a smaller equivalent capacitance. This is because each capacitor 'shares' the voltage. In other words, the total voltage across capacitors is still the same as the source voltage, but each capacitor only has a part of it. Hence, in series, they store less energy compared to when in parallel.

Capacitance in Parallel

When capacitors are connected in parallel, they work together to store charge more effectively. This happens because they all face the same voltage. In this arrangement, the total capacitance is simply the sum of all individual capacitances:\[ C_p = C_1 + C_2 \]In a parallel configuration, capacitors do not divide the voltage among themselves. Instead, they share the same voltage from the source, which allows them to store more energy collectively. This benefit is evident because the equivalent capacitance is larger compared to their series connection. Hence, with the same applied voltage, parallel capacitors store more energy, making this configuration ideal for circuits needing higher energy storage.

Energy Stored in Capacitors

Calculating the energy stored in capacitors is vital for understanding their role in circuits. The energy stored in a capacitor is given by:\[ E = \frac{1}{2} C V^2 \]Where \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. This formula shows that energy storage is dependent on both the capacitance and the voltage. For capacitors in series, with a lower equivalent capacitance, energy storage is less. Conversely, capacitors in parallel, with a higher equivalent capacitance, store more energy. In the exercise, this principle explains why disconnecting and reconnecting capacitors from a series to a parallel configuration results in higher energy storage, making it a more energy-efficient setup for the battery.