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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.

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Most popular questions from this chapter

Which statement about the phase relation between the electric and magnetic fields in an LC circuit is correct? a) When one field is at its maximum, the other is also, and the same for the minimum values. b) When one field is at maximum strength, the other is at minimum (zero) strength. c) The phase relation, in general, depends on the values of \(L\) and \(C\).

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 with 400 turns in its primary coil and 20 turns in its secondary coil is designed to deliver an average power of \(1200 .\) W with a maximum voltage of \(60.0 \mathrm{~V}\). What is the maximum current in the primary coil?

A 10.0 - \(\mu\) F capacitor is fully charged by a 12.0 - \(\mathrm{V}\) battery and is then disconnected from the battery and allowed to discharge through a 0.200 - \(\mathrm{H}\) inductor. Find the first three times when the charge on the capacitor is \(80.0-\mu \mathrm{C},\) taking \(t=0\) as the instant when the capacitor is connected to the inductor.

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