Problem 52 - College Physics

Question

A 4.00\(\mu \mathrm{F}\) and a 6.00\(\mu \mathrm{F}\) capacitor are connected in series, and this combination is connected across a 48.0 \(\mathrm{V}\) potential difference. Calculate (a) the charge on each capacitor and (b) the potential difference across each of them.

Step-by-Step Solution

Answer

Each capacitor has a charge of 115.2 μC. The potential differences are 28.8 V and 19.2 V across the 4.00 μF and 6.00 μF capacitors, respectively.

Step 1: Calculate the equivalent capacitance

When capacitors are connected in series, the equivalent capacitance \( C_{eq} \) can be calculated using the formula: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]Substituting the given values, \( C_1 = 4.00 \, \mu\mathrm{F} \) and \( C_2 = 6.00 \, \mu\mathrm{F} \): \[ \frac{1}{C_{eq}} = \frac{1}{4.00} + \frac{1}{6.00} \]\[ \frac{1}{C_{eq}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \]\[ C_{eq} = \frac{12}{5} = 2.40 \, \mu\mathrm{F} \]

Step 2: Calculate the charge on the equivalent capacitor

To find the charge \( Q \) on the series combination, use the formula:\[ Q = C_{eq} \cdot V \]where \( V = 48.0 \, \mathrm{V} \). Substituting the equivalent capacitance:\[ Q = 2.40 \, \mu\mathrm{F} \times 48.0 \, \mathrm{V} \]\[ Q = 115.2 \, \mu\mathrm{C} \]The charge on each capacitor in series is the same, so each capacitor carries a charge of \( 115.2 \, \mu\mathrm{C} \).

Step 3: Calculate the potential difference across each capacitor

Using the charge calculated in Step 2, find the potential difference \( V_1 \) across the 4.00 \( \mu\mathrm{F} \) capacitor:\[ V_1 = \frac{Q}{C_1} = \frac{115.2 \, \mu\mathrm{C}}{4.00 \, \mu\mathrm{F}} = 28.8 \, \mathrm{V} \]Then, find the potential difference \( V_2 \) across the 6.00 \( \mu\mathrm{F} \) capacitor:\[ V_2 = \frac{Q}{C_2} = \frac{115.2 \, \mu\mathrm{C}}{6.00 \, \mu\mathrm{F}} = 19.2 \, \mathrm{V} \]

Key Concepts

Equivalent Capacitance

When capacitors are connected in series, the overall capacitance behaves differently compared to when they're connected in parallel. To find the equivalent capacitance \( C_{eq} \) for capacitors in series, you use the reciprocal formula:

• \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \)

This formula means that the total capacitance is less than the smallest capacitance in the series. In our example, with a 4.00 \(\mu\mathrm{F}\) and a 6.00 \(\mu\mathrm{F}\) capacitor connected in series:

• The calculation becomes \( \frac{1}{C_{eq}} = \frac{1}{4.00\,\mu\mathrm{F}} + \frac{1}{6.00\,\mu\mathrm{F}} \)
• After combining the fractions, the equivalent capacitance comes out to be \(2.40\,\mu\mathrm{F}\).

This reduced equivalent capacitance occurs because, in a series circuit, each capacitor shares the same charge, leading to an eased overall effect on storing energy compared to capacitors connected in parallel.

Charge on Capacitors

A key aspect of capacitors in series is that they all share the same charge. This is because there is only one path for the charge to flow; hence, the charge on each capacitor is identical.

• To calculate the charge \( Q \) on each capacitor, use the formula: \( Q = C_{eq} \cdot V \)
• Where \( C_{eq} \) is the equivalent capacitance and \( V \) is the applied voltage across the series.

In the given problem, the equivalent capacitance is \(2.40\,\mu\mathrm{F}\) and the voltage is \(48.0\,\mathrm{V}\), leading to:

• \( Q = 2.40\,\mu\mathrm{F} \times 48.0\,\mathrm{V} = 115.2\,\mu\mathrm{C} \)

So, each of the capacitors in the series has a charge of \(115.2\,\mu\mathrm{C}\) on it. This uniformity of charge simplifies the analysis and behavior prediction of series circuits.

Potential Difference Across Capacitors

When dealing with series capacitors, each capacitor can have a different potential difference across it, even though they share the same charge. To find the potential difference across each capacitor, you use:

• \( V = \frac{Q}{C} \)

For our specific capacitors:

• For the 4.00 \(\mu\mathrm{F}\) capacitor: \( V_1 = \frac{115.2\,\mu\mathrm{C}}{4.00\,\mu\mathrm{F}} = 28.8\,\mathrm{V} \)
• For the 6.00 \(\mu\mathrm{F}\) capacitor: \( V_2 = \frac{115.2\,\mu\mathrm{C}}{6.00\,\mu\mathrm{F}} = 19.2\,\mathrm{V} \)

Although the total voltage across the series is \(48.0\,\mathrm{V}\), it's distributed differently across each capacitor. The division depends on individual capacitances, with larger capacitances receiving a smaller share of the total voltage. Understanding this distribution is crucial for ensuring each capacitor's voltage ratings are not exceeded in a circuit design.