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Chapter 21: Problem 4

A circuit breaker trips when the rms current exceeds \(20.0 \mathrm{A} .\) How many \(100.0-\mathrm{W}\) lightbulbs can run on this circuit without tripping the breaker? (The voltage is \(120 \mathrm{V} \mathrm{rms} .)\)

Short Answer

Expert verified
Answer: 24 lightbulbs.

Step by step solution

01

Determine the power consumed by each lightbulb

We are given the power consumed by each lightbulb, which is 100 W.
02

Calculate the current for each lightbulb

To calculate the current for each lightbulb, we will use the equation: Power = Voltage x Current. We are given the voltage (120 V rms) and the power of each lightbulb (100 W), so we can solve for the current (I) using this equation: \(I = \frac{\text{Power}}{\text{Voltage}} = \frac{100 \text{W}}{120 \text{V}} = 0.833 \text{A}\)
03

Determine how many lightbulbs the circuit can support without tripping the breaker

We know that the circuit breaker trips when the rms current exceeds 20.0 A. Since we have calculated the current of each lightbulb to be 0.833 A, we can now determine how many lightbulbs the circuit can support without exceeding the maximum current by dividing the maximum current by the current of each lightbulb: \(\text{Number of lightbulbs} = \frac{\text{Maximum current}}{\text{Current per lightbulb}} = \frac{20.0 \text{A}}{0.833 \text{A}} = 24\) So 24 lightbulbs (rounded down) can be run on this circuit without tripping the breaker.

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Most popular questions from this chapter

(a) What mas current is drawn by a \(4200-\mathrm{W}\) electric room heater when running on \(120 \mathrm{V}\) rms? (b) What is the power dissipation by the heater if the voltage drops to \(105 \mathrm{V}\) rms during a brownout? Assume the resistance stays the same.
The instantaneous sinusoidal emf from an ac generator with an mos emf of $4.0 \mathrm{V}$ oscillates between what values?
For a particular \(R L C\) series circuit, the capacitive reactance is $12.0 \Omega,\( the inductive reactance is \)23.0 \Omega,$ and the maximum voltage across the \(25.0-\Omega\) resistor is \(8.00 \mathrm{V}\) (a) What is the impedance of the circuit? (b) What is the maximum voltage across this circuit?
A \(150-\Omega\) resistor is in series with a \(0.75-\mathrm{H}\) inductor in an ac circuit. The rms voltages across the two are the same. (a) What is the frequency? (b) Would each of the rms voltages be half of the rms voltage of the source? If not, what fraction of the source voltage are they? (In other words, \(V_{R} / \ell_{m}=V_{L} / \mathcal{E}_{m}=?\) ) (c) What is the phase angle between the source voltage and the current? Which leads? (d) What is the impedance of the circuit?
The circuit shown has a source voltage of \(440 \mathrm{V}\) ms, resistance \(R=250 \Omega\), inductance \(L=0.800 \mathrm{H},\) and capacitance $C=2.22 \mu \mathrm{F} .\( (a) Find the angular frequency \)\omega_{0}$ for resonance in this circuit. (b) Draw a phasor diagram for the circuit at resonance. (c) Find these rms voltages measured between various points in the circuit: \(V_{a b}, V_{b c}, V_{c d}, V_{b d},\) and \(V_{c d},\) (d) The resistor is replaced with one of \(R=125 \Omega\). Now what is the angular frequency for resonance? (e) What is the rms current in the circuit operated at resonance with the new resistor?
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