Problem 18 - Physics

Question

A light bulb has a resistance of \(240 \Omega\). It is connected to a standard wall socket (120 V, \(60.0 \mathrm{~Hz}\) ). (a) Determine the current in the bulb. (b) Determine the current in the bulb after a \(10.0-\mu \mathrm{F}\) capacitor is added in series in the circuit. (c) It is possible to return the current in the bulb to the value calculated in part (a) by adding an inductor in series with the bulb and the capacitor. What is the value of the inductance of this inductor?

Step-by-Step Solution

Answer

(a) 0.5 A (b) 0.335 A (c) 0.704 H

Step 1: Understanding Ohm's Law

To find the initial current through the light bulb, we can use Ohm's Law: \( I = \frac{V}{R} \). Here, \( V = 120 \text{ V} \) and \( R = 240 \text{ Ω} \).

Step 2: Calculating Initial Current

Using Ohm's Law, substitute the given values into the formula: \[ I_a = \frac{120 \text{ V}}{240 \text{ Ω}} = 0.5 \text{ A} \]. Thus, the initial current is 0.5 A.

Step 3: Calculating the Total Impedance with the Capacitor

Once the capacitor is added in series, the impedance of the circuit changes. For a capacitive reactance, the formula is \( X_C = \frac{1}{2 \pi f C} \) where \( f = 60 \text{ Hz} \) and \( C = 10.0 \times 10^{-6} \text{ F} \).

Step 4: Determining Capacitive Reactance

Using \( X_C = \frac{1}{2 \pi \times 60 \times 10.0 \times 10^{-6}} \), calculate \( X_C \): \[ X_C \approx 265.26 \text{ Ω} \].

Step 5: Calculating New Impedance and Current

In a series circuit, total impedance \( Z \) is given by \( Z = \sqrt{R^2 + X_C^2} \). Thus, \( Z = \sqrt{240^2 + 265.26^2} \approx 357.77 \text{ Ω} \). Then, calculate the new current \( I_b \): \[ I_b = \frac{120}{357.77} \approx 0.335 \text{ A} \].

Step 6: Determine Inductor's Reactance to Match Initial Current

To bring the current back to its initial value, the circuit must be adjusted such that \( |Z_{RL}| = 240 \text{ Ω} \). The inductive reactance \( X_L \) must satisfy \( Z = \sqrt{R^2 + (X_L - X_C)^2} = 240 \).

Step 7: Solving for Inductive Reactance

Solve \( 240 = \sqrt{240^2 + (X_L - 265.26)^2} \), which simplifies to \( X_L - 265.26 = 0 \). Thus, \( X_L = 265.26 \text{ Ω} \).

Step 8: Calculating the Required Inductance

Since \( X_L = 2 \pi f L \), rearrange to solve for \( L \): \( L = \frac{X_L}{2 \pi f} \). Compute \( L = \frac{265.26}{2 \pi \times 60} \approx 0.704 \text{ H} \).

Key Concepts

Capacitance

Capacitance is a measure of a capacitor's ability to store electrical charge. It is denoted by the symbol \( C \) and its unit is the farad (F). Capacitors play a key role in many electronic circuits by introducing a phase shift and reacting to changing frequencies.
A capacitor, when added in series to a circuit, affects the circuit's impedance. The effect can be quantified using the capacitive reactance, denoted as \( X_C \). The capacitive reactance is inversely proportional to both the frequency \( f \) of the applied AC signal and the capacitance \( C \) itself. The formula for capacitive reactance is:

• \( X_C = \frac{1}{2 \pi f C} \)

This means that higher frequencies or larger capacitances lead to lower reactance, allowing more current to pass through the circuit. Understanding these interactions is essential for analyzing AC circuits and troubleshooting issues that involve capacitors.

Inductance

Inductance is a property of an electrical conductor that opposes changes in current. It is represented by the symbol \( L \) and its unit is the henry (H). Inductors are components that typically consist of a coil of wire and are used to induce a magnetic field when electrical current passes through them.
In AC circuits, inductors introduce inductive reactance, denoted as \( X_L \), which opposes changes in current flow through the circuit. This inductive reactance can be expressed by the formula:

• \( X_L = 2 \pi f L \)

Where \( f \) is the frequency of the AC signal.
In the given exercise, inductance is practically employed to adjust current levels in the circuit. By carefully selecting an inductor with the proper value of \( L \), it's possible to counteract the effect introduced by a series capacitor. In the example, achieving a total circuit impedance equal to the original resistance ensures the initial current value is restored. This demonstrates how inductors can be used effectively in tuning and balancing circuits.

Impedance

Impedance is a broad measure of opposition in an AC circuit, comprising both the resistance (from resistors) and reactance (from capacitors and inductors). It is denoted as \( Z \) and its unit is the ohm (Ω). Impedance encompasses the combined effect of all these elements within a circuit.
The presence of capacitive and inductive elements introduces reactance, which can be either capacitive or inductive. The total impedance \( Z \) in a series circuit involving resistance \( R \), capacitive reactance \( X_C \), and inductive reactance \( X_L \) is given by:

• \( Z = \sqrt{R^2 + (X_L - X_C)^2} \)

This formula illustrates how both the magnitude and direction of the reactance affect the total impedance. Particularly, when capacitors and inductors are in series, their reactances partially cancel each other out.
In practical terms, impedance helps in calculating how a circuit behaves with AC signals in terms of current and voltage relationships. By understanding and calculating impedance, you can predict and manage how different components in your circuit influence current flow and power distribution.