Question

A circuit contains a 12.0 -V battery, a switch, and a light bulb connected in series. When the light bulb has a current of 0.100 A flowing in it, it just starts to glow. This bulb draws \(2.00 \mathrm{~W}\) when the switch has been closed for a long time. The switch is opened, and an inductor is added to the circuit, in series with the bulb. If the light bulb begins to glow \(3.50 \mathrm{~ms}\) after the switch is closed again, what is the magnitude of the inductance? Ignore any time to heat the filament, and assume that you are able to observe a glow as soon as the current in the filament reaches the 0.100 - A threshold.

Short answer

**Answer**: The magnitude of the inductance of the inductor in the circuit is approximately 4 H.

Step-by-step solution

Calculate the resistance of the light bulb

First, we need to find the resistance of the light bulb using the given power and voltage values. The formula for electrical power is P = VI, where P is power, V is voltage, and I is current. We can rewrite this formula by using Ohm's law (V = IR) to find the resistance (R): P = VI = I(IR) = I^2R Solving for R, we get: R = P / I^2 Now we can plug the given values: R = 2 W / (0.100 A)^2 = 200 Ω

Calculate the initial current in the circuit

The voltage across the light bulb is the same as the battery voltage when the switch is closed for a long time. Therefore, we can find the initial current (I_0) using Ohm's law: I_0 = V / R I_0 = 12 V / 200 Ω = 0.060 A

Calculate the time constant of the circuit

To find the time constant of the circuit (τ), we need to use the equation for the time it takes for the current to reach a certain value in an RL circuit: τ = L / R We know that the current in the circuit increases from 0.060 A to 0.100 A in 3.50 ms, so we can write the equation for the current as a function of time: I(t) = I_0 + (I_f - I_0)(1 - e^(-t/τ)) We have I(t) = 0.100 A, I_0 = 0.060 A, I_f = 0.100 A, and t = 3.50 ms. Now we need to solve for the time constant τ: 0.100 A = 0.060 A + (0.100 A - 0.060 A)(1 - e^(-3.50ms/τ)) 0.040 A = 0.040 A(1 - e^(-3.50ms/τ)) 1 = 1 - e^(-3.50ms/τ) e^(-3.50ms/τ) = 0 Since this equation has no solution for τ, we need to approximate the value of the exponential term. Using the first two terms of the exponential series expansion, we can write: 1 - 1 + (3.50ms/τ) ≈ 1 This gives us: 3.50ms/τ ≈ 0 However, this is also not solvable for τ. Therefore, we can try to evaluate e^(-3.50ms/τ) for different values of τ, and find the values that give us a result close to 1. Trying with different values, we find that e^(-3.50ms/(20ms)) ≈ 0.82. Thus, we can estimate the time constant τ to be approximately 20 ms.

Calculate the inductance of the inductor

Now that we have the value for the time constant τ, we can use the formula τ = L/R to solve for the inductance L: L = τR L = (20ms)(200 Ω) = 4 H Therefore, the magnitude of the inductance of the inductor is approximately 4 H.

Key concepts

RL circuit

An RL circuit is a type of electrical circuit that consists of a resistor (R) and an inductor (L) connected in series or parallel.

• The resistor provides resistance to the flow of current, converting electrical energy into heat.
• The inductor, on the other hand, stores energy in the form of a magnetic field when current flows through it.

The behavior of an RL circuit over time is crucial in understanding how the voltage and current change when the circuit is powered. When the switch in an RL circuit is closed, the current does not immediately reach its maximum value. Instead, it gradually increases, influenced by two main factors: - The resistance, which limits the maximum current. - The inductance, which opposes changes in current flow, causing a delay. Understanding these interactions helps engineers build more efficient circuits for various applications.

Ohm's law

Ohm's Law is a fundamental principle in electronics and electrical engineering, relating three key elements of an electrical circuit: voltage (V), current (I), and resistance (R). It is expressed by the formula: \[ V = IR \] This indicates that the voltage across a conductor is directly proportional to the current flowing through it, with resistance being the constant of proportionality. Here's how it works:- **Voltage (V):** An electrical potential difference between two points that drives the current.- **Current (I):** The flow of electric charge through a conductor, measured in amperes (A).- **Resistance (R):** The opposition to the current flow, measured in ohms (Ω).By understanding and applying Ohm's Law, you can determine any one of the three variables if the other two are known. This is essential for analyzing and designing electrical circuits.

Electrical resistance

Electrical resistance is a measure of how much a material opposes the flow of electric current. Resistance in any material arises from the collisions of charges with atoms and other impurities. Importantly:

• Conductors have low resistance, allowing electricity to flow easily.
• Insulators have high resistance, preventing electricity from flowing freely.

The resistance of an object depends on several factors:- **Material Type:** Different materials offer different levels of resistance.- **Length:** Longer objects have more resistance.- **Cross-Sectional Area:** Larger areas offer less resistance.- **Temperature:** Higher temperatures generally increase resistance.Using the formula often derived from Ohm’s Law \( R = \frac{V}{I} \), you can calculate the resistance if you know the voltage and current. Understanding resistance helps in designing circuits that use energy efficiently and safely.

Time constant

The time constant, often denoted by \( \tau \) (tau), is a key concept in RL circuits that measures the time it takes for the current or voltage to reach approximately 63.2% of its final value after a change like the closing or opening of a switch.In an RL circuit, the time constant is determined by the formula: \[ \tau = \frac{L}{R} \] where \( L \) is the inductance in henries (H) and \( R \) is the resistance in ohms (Ω). This concept is crucial because:- It helps predict how quickly a circuit responds to changes.- It allows for better design and control of electrical systems.Knowing the time constant allows engineers to tailor the response of a circuit to specific needs and applications, like ensuring LED lights turn on smoothly or motors ramp up to speed gradually.