Problem 5 - Physics

Question

A circuit consists of a \(3.00-\mu F\) and a \(6.00-\mu F\) capacitor connected in series across the terminals of a 510 -Hz generator. The voltage of the generator is \(120 \mathrm{~V}\). (a) Determine the equivalent capacitance of the two capacitors. (b) Find the current in the circuit.

Step-by-Step Solution

Answer

(a) The equivalent capacitance is 2.00 µF. (b) The current in the circuit is approximately 0.774 A.

Step 1: Calculate Equivalent Capacitance

For capacitors in series, the equivalent capacitance \( C_{eq} \) is determined using the formula: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \] where \( C_1 = 3.00\,\mu F \) and \( C_2 = 6.00\,\mu F \). Substituting the values, we have: \[ \frac{1}{C_{eq}} = \frac{1}{3.00} + \frac{1}{6.00} \] Simplifying, \[ \frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \] Therefore, \( C_{eq} = 2.00\,\mu F \).

Step 2: Calculate Capacitive Reactance

The capacitive reactance \( X_C \) is calculated using the formula: \[ X_C = \frac{1}{2\pi f C_{eq}} \] where \( f = 510\,\text{Hz} \) and \( C_{eq} = 2.00\,\mu F = 2.00 \times 10^{-6}\,F \). Substituting these values, \[ X_C = \frac{1}{2\pi \times 510 \times 2.00 \times 10^{-6}} \] Simplifying, \[ X_C \approx 155.04\,\Omega \].

Step 3: Calculate Current in the Circuit

The current \( I \) in the circuit can be calculated using Ohm's Law for AC circuits, \( I = \frac{V}{X_C} \) where \( V = 120\,V \). Using the value of \( X_C \approx 155.04\,\Omega \) calculated earlier, \[ I = \frac{120}{155.04} \approx 0.774\,A \].

Key Concepts

Equivalent Capacitance

When we join capacitors in series, they behave differently compared to when they are connected in parallel. In a series combination, the total capacitance decreases, and the equivalent capacitance can be calculated using the formula:

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

This formula shows that we are essentially adding the reciprocals of each capacitance. For instance, in our exercise with a 3.00-µF and a 6.00-µF capacitor, we can calculate the equivalent capacitance as:

• \( \frac{1}{C_{eq}} = \frac{1}{3.00} + \frac{1}{6.00} \)

Upon simplifying, we find that the equivalent capacitance, \( C_{eq} \), is 2.00 µF. Understanding this concept helps in designing circuits based on specific capacitance requirements.
By lowering the total capacitance in series, charges storing capacity shifts and circuit behavior changes accordingly.

Capacitive Reactance

Capacitive reactance represents a capacitor's opposition to the flow of alternating current (AC). It is similar to resistance in a direct current (DC) circuit but varies with frequency. The relationship is given by:

• \( X_C = \frac{1}{2\pi f C_{eq}} \)

In practice, as the frequency \( f \) rises, or the capacitance \( C_{eq} \) decreases, the capacitive reactance \( X_C \) diminishes, meaning the circuit allows more current to pass. Using our given frequency of 510 Hz and an equivalent capacitance of 2.00 µF, we computed:

• \( X_C \approx 155.04\,\Omega \)

This calculation shows how capacitors can be used to control current flow in AC circuits by adjusting frequency or capacitance. This is particularly useful in radio, audio applications, and filters.
Learning to calculate capacitive reactance is vital for grasping how AC circuits function.

Ohm's Law for AC Circuits

The behavior of current in AC circuits is explained through an adaptation of Ohm's Law, which relates voltage, current, and reactance (instead of resistance):

• \( I = \frac{V}{X_C} \)

This form is crucial because it highlights that reactance—reactive components like capacitors and inductors—affects the circuit like resistance does in DC circuits. For example, in our problem, with a generator voltage of 120 V and a capacitive reactance of approximately 155.04 \( \Omega \), the current \( I \) can be computed:

• \( I \approx 0.774\,A \)

This exemplifies the usefulness of Ohm's Law in AC circuits for predicting behavior and managing power consumption, aiding in the design and analysis of complex AC systems like power distribution and AC motor operations.
By understanding and using this law, you're prepared for a variety of practical applications in electrical engineering.