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Chapter 30: Problem 48

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

A transformer contains a primary coil with 200 turns and a secondary coil with 120 turns. The secondary coil drives a current \(I\) through a \(1.00-\mathrm{k} \Omega\) resistor. If an input voltage \(V_{\mathrm{rms}}=75.0 \mathrm{~V}\) is applied across the primary coil, what is the power dissipated in the resistor?

The AM radio band covers the frequency range from \(520 \mathrm{kHz}\) to \(1610 \mathrm{kHz}\). Assuming a fixed inductance in a simple LC circuit, what ratio of capacitance is necessary to cover this frequency range? That is, what is the value of \(C_{\mathrm{h}} / C_{\mathrm{l}}\) where \(C_{\mathrm{h}}\) is the capacitance for the highest frequency and \(C_{1}\) is the capacitance for the lowest frequency? a) 9.59 b) 0.104 c) 0.568 d) 1.76

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 capacitor with capacitance \(C=5.00 \cdot 10^{-6} \mathrm{~F}\) is connected to an AC power source having a peak value of \(10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Find the reactance of the capacitor and the maximum current in the circuit.

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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