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Chapter 21: Problem 3

A \(1500-\) W heater runs on \(120 \mathrm{V}\) ms. What is the peak current through the heater? (Wtorial: power in ac circuits)

Short Answer

Expert verified
Answer: The peak current through the heater is approximately 17.68 A.

Step by step solution

01

Determine the rms current

To find the rms current through the heater, use the power formula for AC circuits: $$ P = V_{rms} \cdot I_{rms} $$ Rearrange the formula to solve for rms current: $$ I_{rms} = \frac{P}{V_{rms}} $$ Substitute the given values for power and voltage: $$ I_{rms} = \frac{1500 \mathrm{W}}{120 \mathrm{V}} $$
02

Calculate the rms current

Divide the power by the voltage to find the rms current: $$ I_{rms} = \frac{1500 \mathrm{W}}{120 \mathrm{V}} = 12.5 \ \mathrm{A} $$ So, the rms current through the heater is 12.5 A.
03

Determine the peak current

Since we have the rms current, we can now find the peak current using the relationship between the rms and peak currents: $$ I_{peak} = \sqrt{2} \cdot I_{rms} $$ Substitute the given rms current value: $$ I_{peak} = \sqrt{2} \cdot 12.5 \ \mathrm{A} $$
04

Calculate the peak current

Multiply the rms current by the square root of 2 to find the peak current: $$ I_{peak} = \sqrt{2} \cdot 12.5 \ \mathrm{A} \approx 17.68 \ \mathrm{A} $$ The peak current through the heater is approximately 17.68 A.

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Most popular questions from this chapter

A 25.0 -mH inductor, with internal resistance of \(25.0 \Omega\) is connected to a \(110-\mathrm{V}\) ms source. If the average power dissipated in the circuit is \(50.0 \mathrm{W},\) what is the frequency? (Model the inductor as an ideal inductor in series with a resistor.)
An alternator supplics a peak current of \(4.68 \mathrm{A}\) to a coil with a negligibly small internal resistance. The voltage of the alternator is \(420-\mathrm{V}\) peak at \(60.0 \mathrm{Hz}\) When a capacitor of $38.0 \mu \mathrm{F}$ is placed in series with the coil, the power factor is found to be \(1.00 .\) Find (a) the inductive reactance of the coil and (b) the inductance of the coil.
An \(R L C\) series circuit has a resistance of \(R=325 \Omega\) an inductance \(\quad L=0.300 \mathrm{mH}, \quad\) and \(\quad\) a capacitance $C=33.0 \mathrm{nF} .$ (a) What is the resonant frequency? (b) If the capacitor breaks down for peak voltages in excess of \(7.0 \times 10^{2} \mathrm{V},\) what is the maximum source voltage amplitude when the circuit is operated at the resonant frequency?
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?
An ac circuit contains a \(12.5-\Omega\) resistor, a \(5.00-\mu \mathrm{F}\) capacitor, and a \(3.60-\mathrm{mH}\) inductor connected in series to an ac generator with an output voltage of \(50.0 \mathrm{V}\) (peak) and frequency of \(1.59 \mathrm{kHz}\). Find the impedance,
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