Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 30: Problem 70

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

Expert verified
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

01

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: \[S = 200 A \times 440 V = 88000 W\]
02

Calculate Real Power

The real power (P) of the light bulb is given in the problem as 75000 W.
03

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: \[PF = \frac{75000}{88000} = 0.85227\]
04

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.
05

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: \[Z = \frac{440 V}{200 A} = 2.2 \Omega\]
06

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: \[X_{L} = \sqrt{(2.2 \Omega)^{2} - (1.62 \Omega)^{2}} = \sqrt{4.84 - 2.6244} = \sqrt{2.2156} = 1.488 \Omega\]
07

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An inductor with inductance \(L=47.0 \mathrm{mH}\) is connected to an AC power source having a peak value of \(12.0 \mathrm{~V}\) and \(f=1000 . \mathrm{Hz} .\) Find the reactance of the inductor and the maximum current in the circuit.

If you use a parallel plate capacitor with air in the gap between the plates as part of a series RLC circuit in a generator, you can measure current flowing through the generator. Why is it that the air gap in the capacitor does not act like an open switch, blocking all current flow in the circuit?

A 2.00 - \(\mu\) F capacitor was fully charged by being connected to a 12.0 - \(V\) battery. The fully charged capacitor is then connected in series with a resistor and an inductor: \(R=50.0 \Omega\) and \(L=0.200 \mathrm{H}\). Calculate the damped frequency of the resulting circuit.

A transformer has 800 turns in the primary coil and 40 turns in the secondary coil. a) What happens if an AC voltage of \(100 . \mathrm{V}\) is across the primary coil? b) If the initial \(A C\) current is \(5.00 \mathrm{~A}\), what is the output current? c) What happens if a DC current at \(100 .\) V flows into the primary coil? d) If the initial DC current is \(5.00 \mathrm{~A}\), what is the output current?

30.24 A 2.00 - \(\mu\) F capacitor is fully charged by being connected to a 12.0 - \(\mathrm{V}\) battery. The fully charged capacitor is then connected to a \(0.250-\mathrm{H}\) inductor. Calculate (a) the maximum current in the inductor and (b) the frequency of oscillation of the LC circuit.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Quantum Physics

Read Explanation

Rotational Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Waves Physics

Read Explanation

Linear Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free