Question

A 75,000 -W light bulb (yes, there are such things!) operates at \(I_{\mathrm{rms}}=200 . \mathrm{A}\) and \(V_{\mathrm{rms}}=440 . \mathrm{V}\) in a \(60.0-\mathrm{Hz} \mathrm{AC}\) circuit. Find the resistance, \(R\), and self- inductance, \(L\), of this bulb. Its capacitive reactance is negligible.

Short answer

Question: Determine the resistance (R) and self-inductance (L) of a light bulb in an AC circuit with a power of 75000 W, rms current of 200 A, rms voltage of 440 V, and frequency of 60 Hz. Answer: The resistance (R) of the light bulb is approximately 1.62 Ω, and the self-inductance (L) is approximately 0.00395 H.

Step-by-step solution

Calculate Apparent Power

First, we need to calculate the apparent power (S) of the light bulb, using the rms current (I_rms) and rms voltage (V_rms). The formula for apparent power is: \[S = I_{\mathrm{rms}} \times V_{\mathrm{rms}}\] Substitute the given values: \[S = 200 A \times 440 V = 88000 W\]

Calculate Real Power

The real power (P) of the light bulb is given in the problem as 75000 W.

Calculate Power Factor

The power factor (PF) is the ratio of real power to apparent power. The formula for the power factor is: \[PF = \frac{P}{S}\] Substituting the known values: \[PF = \frac{75000}{88000} = 0.85227\]

Calculate Resistance

Using the power factor and the rms voltage and current, we can calculate the resistance (R) in the circuit. The formula is: \[R = \frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}} \times PF\] Plugging in the values: \[R = \frac{440 V}{200 A} \times 0.85227 = 1.902 V/A \times 0.85227 = 1.62 \Omega\] So, the resistance of the light bulb is approximately 1.62 Ω.

Calculate Impedance

Using Ohm's law for AC circuits, we can find the impedance (Z) of the circuit. The formula for impedance is: \[Z = \frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}}\] Substitute the known values: \[Z = \frac{440 V}{200 A} = 2.2 \Omega\]

Calculate Inductive Reactance

Since we now know the impedance and resistance, we can calculate the inductive reactance (X_L) using the following formula: \[X_{L} = \sqrt{Z^{2} - R^{2}}\] Plugging in the values: \[X_{L} = \sqrt{(2.2 \Omega)^{2} - (1.62 \Omega)^{2}} = \sqrt{4.84 - 2.6244} = \sqrt{2.2156} = 1.488 \Omega\]

Calculate Self-Inductance

Finally, we can calculate the self-inductance (L) using the formula for inductive reactance and the given frequency (f) in the circuit: \[X_{L} = 2 \pi f L\] Solving for L, we have: \[L = \frac{X_{L}}{2 \pi f}\] Substitute the values: \[L = \frac{1.488 \Omega}{2 \pi \times 60 Hz} = \frac{1.488 \Omega}{377} = 0.00395 H\] So, the self-inductance of the light bulb is approximately 0.00395 H.

Key concepts

Apparent Power

In an AC circuit like the one described for the high-power light bulb, apparent power is a key concept. It is represented by the variable \( S \) and signifies the product of the root mean square (RMS) voltage \( V_{\text{rms}} \) and RMS current \( I_{\text{rms}} \). Apparent power essentially quantifies the total power flowing through an AC circuit, without distinguishing how much of it is actually used to perform work.
For our exercise, the apparent power is calculated with this formula:

• \( S = I_{\text{rms}} \times V_{\text{rms}} \)

By inserting the known values, \( I_{\text{rms}} = 200 \ A \) and \( V_{\text{rms}} = 440 \ V \), we find \( S = 88000 \ W \).
This results in an apparent power of 88000 watts, meaning this is the total power present in the circuit. This number makes it clear why distinguishing between different types of power in AC circuits is crucial, as apparent power includes both power used for actual work and power that oscillates back and forth in the system.

Power Factor

The power factor is crucial in determining how efficiently the electrical power is being used within the AC circuit. It is defined as the ratio of real power, the power that does actual work, to apparent power, which includes all power in the circuit.
Its value ranges between 0 and 1, with higher values indicating more efficient power usage.
The formula for calculating the power factor \( PF \) is:

• \( PF = \frac{P}{S} \)

Given that the real power \( P \) of our light bulb is 75000 watts and the apparent power \( S \) is 88000 watts, we can find that the power factor is approximately \( 0.85227 \).
This indicates that around 85% of the power is used for actual work, while the remaining portion is reactive. Understanding the power factor helps in optimizing energy consumption and minimizing loss in electrical systems.

Impedance

In AC circuits, impedance is a comprehensive measure of opposition against the flow of electric current, combining both resistance and reactance. It is denoted by \( Z \) and is essential for understanding how different components in the circuit affect the current flow.
Using Ohm’s Law in AC circuit form, impedance can be calculated with:

• \( Z = \frac{V_{\text{rms}}}{I_{\text{rms}}} \)

In our light bulb scenario, with an RMS voltage \( V_{\text{rms}} = 440 \ V\) and an RMS current \( I_{\text{rms}} = 200 \ A \), we calculate the impedance as \( Z = 2.2 \ \Omega \).
This means the total opposition to the current due to both resistance and reactance is 2.2 ohms. Impedance helps us understand the overall behavior of circuits in terms of power flow and loss.

Inductive Reactance

Inductive reactance is a specific aspect of reactance, related to the opposition of current change in inductors. It arises in AC circuits because inductors resist changes in current, resulting in a phase difference between voltage and current. This phase difference contributes to apparent power but does not contribute to real power.
Inductive reactance \( X_L \) can be determined from impedance \( Z \) and resistance \( R \), using:

• \( X_L = \sqrt{Z^2 - R^2} \)

For the given circuit, where \( Z = 2.2 \ \Omega \) and \( R = 1.62 \ \Omega \), we find \( X_L \) to be approximately \( 1.488 \ \Omega \).
Additionally, inductive reactance is related to frequency \( f \) and inductance \( L \) with the formula:

• \( X_L = 2 \pi f L \)

Solving for \( L \), we get a self-inductance of approximately \( 0.00395 \ H \), which shows the inductor's ability to store energy in a magnetic field. Analyzing inductive reactance is key to designing efficient circuits and proper energy management.