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A label on a hair dryer reads "110V \(1250 \mathrm{~W}\)." What is the peak current in the hair dryer, assuming that it behaves like a resistor?

Short Answer

Expert verified
Answer: The peak current in the hair dryer is 16.06 A.

Step by step solution

01

Find effective (root mean square) current

Using the power equation, we can find the effective (RMS) current: \(P = IV\) \(I_{rms} = \frac{P}{V}\) Plug in the given values for power and voltage: \(I_{rms} = \frac{1250 \mathrm{~W}}{110 \mathrm{~V}}\) Calculate the RMS current: \(I_{rms} = 11.36 \mathrm{~A}\)
02

Find the resistance

Now, we will use Ohm's Law to find the resistance of the hair dryer: \(V = IR\) \(R = \frac{V}{I}\) Plug in the given value for voltage and the calculated effective current: \(R = \frac{110 \mathrm{~V}}{11.36 \mathrm{~A}}\) Calculate the resistance: \(R = 9.68 \mathrm{~\Omega}\)
03

Find the peak current

Finally, we will use the peak current equation to find the peak current in the hair dryer: \(I_{peak} = I_{rms} \sqrt{2}\) Plug in the calculated RMS current: \(I_{peak} = 11.36 \mathrm{~A} \sqrt{2}\) Calculate the peak current: \(I_{peak} = 16.06 \mathrm{~A}\) So, the peak current in the hair dryer is \(16.06 \mathrm{~A}\).

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Most popular questions from this chapter

The figure shows a simple FM antenna circuit in which \(L=8.22 \mu \mathrm{H}\) and \(C\) is variable (the capacitor can be tuned to receive a specific station). The radio signal from your favorite FM station produces a sinusoidal time-varying emf with an amplitude of \(12.9 \mu \mathrm{V}\) and a frequency of \(88.7 \mathrm{MHz}\) in the antenna. a) To what value, \(C_{0}\), should you tune the capacitor in order to best receive this station? b) Another radio station's signal produces a sinusoidal time-varying emf with the same amplitude, \(12.9 \mu \mathrm{V}\), but with a frequency of \(88.5 \mathrm{MHz}\) in the antenna. With the circuit tuned to optimize reception at \(88.7 \mathrm{MHz}\), what should the value, \(R_{0}\), of the resistance be in order to reduce by a factor of 2 (compared to the current if the circuit were optimized for \(88.5 \mathrm{MHz}\) ) the current produced by the signal from this station?

A variable capacitor used in an RLC circuit produces a resonant frequency of \(5.0 \mathrm{MHz}\) when its capacitance is set to \(15 \mathrm{pF}\). What will the resonant frequency be when the capacitance is increased to \(380 \mathrm{pF} ?\)

a) A loop of wire \(5.00 \mathrm{~cm}\) in diameter is carrying a current of \(2.00 \mathrm{~A}\). What is the energy density of the magnetic field at its center? b) What current has to flow in a straight wire to produce the same energy density at a point \(4.00 \mathrm{~cm}\) from the wire?

The transmission of electric power occurs at the highest possible voltage to reduce losses. By how much could the power loss be reduced by raising the voltage by a factor of \(10 ?\)

A series circuit contains a \(100.0-\Omega\) resistor, a \(0.500-\mathrm{H}\) inductor, a 0.400 - \(\mu\) F capacitor, and a time-varying source of emf providing \(40.0 \mathrm{~V}\). a) What is the resonant angular frequency of the circuit? b) What current will flow through the circuit at the resonant frequency?

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