Problem 86 - Physics

Question

Three capacitors \((4.0,6.0\), and \(12.0 \mu \mathrm{F})\) are connected in series across a \(50.0-\mathrm{V}\) battery. Find the voltage across the \(4.0-\mu \mathrm{F}\) capacitor.

Step-by-Step Solution

Answer

The voltage across the 4.0 μF capacitor is 25.0 V.

Step 1: Start with the Formula for Capacitors in Series

Capacitors in series follow the formula for the equivalent capacitance: \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \), where \(C_1\), \(C_2\), and \(C_3\) are the capacitances of the individual capacitors.

Step 2: Calculate the Equivalent Capacitance

Substitute the capacitance values into the formula: \( \frac{1}{C_{eq}} = \frac{1}{4.0} + \frac{1}{6.0} + \frac{1}{12.0} \). This gives \( \frac{1}{C_{eq}} = \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2} \). So, \( C_{eq} = 2.0 \mu F \).

Step 3: Use Charge to Determine Voltage Across Each Capacitor

In series, the charge \(Q\) on each capacitor is the same. Calculate \(Q\) using \( Q = C_{eq} \times V_{total} = 2.0 \mu F \times 50.0 V = 100.0 \mu C \).

Step 4: Calculate Voltage Across the 4.0 μF Capacitor

The voltage across a single capacitor in series is given by \( V = \frac{Q}{C} \). Thus, \( V_{4.0 \mu F} = \frac{100.0 \mu C}{4.0 \mu F} = 25.0 V \).

Key Concepts

Equivalent Capacitance

When capacitors are connected in series, their total effect or combined capacitance becomes different from just adding them up. This combined measurement is known as the equivalent capacitance. The formula for calculating this is:

• \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \)

where \( C_1 \), \( C_2 \), and \( C_3 \) are the individual capacitances. It might seem tricky because you need to take the reciprocal of each capacitance first. After finding the sum of those, you get the reciprocal of the total, \( C_{eq} \). For instance, with capacitances of \( 4.0 \ \mu F \), \( 6.0 \ \mu F \), and \( 12.0 \ \mu F \), we calculated the equivalent capacitance as \( 2.0 \ \mu F \). Remember, individual capacitances in series always result in a lower equivalent capacitance than any given individual capacitance.

Voltage Across Capacitor

Once you know the equivalent capacitance and the total voltage across a series of capacitors, you can find out how this voltage is distributed across each capacitor. In a series circuit, the capacitors share the total voltage according to their capacitance values. The voltage across each capacitor can be found using the charge \( Q \) that is common to each in the series and the formula:

• \( V = \frac{Q}{C} \)

For the \( 4.0 \ \mu F \) capacitor in our example, the voltage across it was calculated using the charge \( 100.0 \ \mu C \) determined from the equivalent capacitance, resulting in a voltage of \( 25.0 \ V \). Each capacitor has a higher or lower voltage based on its capacitance, sharing the total \( 50 \ V \) supplied by the battery across the series line.

Charge Calculation

Understanding charge calculation in a series capacitor circuit is crucial as it affects voltage calculations. In such circuits, the charge \( Q \) remains consistent across all capacitors. This means every capacitor, regardless of its capacitance, stores the same amount of electrical charge. Calculating charge involves using the equivalent total capacitance and the total voltage applied across the series setup:

• \( Q = C_{eq} \times V_{total} \)

For our scenario, with a \( C_{eq} \) of \( 2.0 \ \mu F \) and a battery voltage of \( 50.0 \ V \), the charge on each capacitor was computed as \( 100.0 \ \mu C \). This uniform charge explains why different capacitors will each have different voltages, based on their size, but hold the same charge. Understanding this consistent share of charge helps in predicting circuit behavior.