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Chapter 30: Problem 61

What are the maximum values of (a) current and (b) voltage when an incandescent 60 -W light bulb (at \(110 \mathrm{~V})\) is connected to a wall plug labeled \(110 \mathrm{~V} ?\)

Short Answer

Expert verified
Answer: The maximum current for the light bulb is calculated using the formula \(I_\text{max} = \frac{110 \mathrm{~V}}{R}\), where R is the resistance of the light bulb. The maximum voltage for the light bulb is given and is 110 V.

Step by step solution

01

Calculate the resistance of the light bulb

First, we need to find the resistance of the light bulb. We can do this using the power formula: \(P = V^2/R\), where P is the power in watts, V is the voltage in volts, and R is the resistance in ohms. Rearranging the formula for R gives: \(R = \frac{V^2}{P}\). Given the power of the bulb (P) to be 60 W and the voltage (V) to be 110 V, we can plug in these values and calculate the resistance of the light bulb: \(R = \frac{(110 \mathrm{~V})^2}{60 \mathrm{~W}}\)
02

Calculate the maximum current

Now that we have the resistance, we can use Ohm's law to calculate the maximum current. Ohm's law states: \(I = \frac{V}{R}\), where I is the current in amperes, V is the voltage in volts, and R is the resistance in ohms. Using the voltage given for the wall plug, 110 V, and the calculated resistance, we can find the maximum current: \(I_\text{max} = \frac{110 \mathrm{~V}}{R}\)
03

Calculate the maximum voltage

The maximum voltage is already given by the problem as 110 V. This is because the voltage across the light bulb is determined by the voltage of the wall plug to which it is connected. So, \(V_\text{max} = 110 \mathrm{~V}\)
04

Final answers

Calculating the values from the previous steps, we find the maximum current and voltage for the light bulb: (a) Maximum Current: \(I_\text{max} = \frac{110 \mathrm{~V}}{R}\) (b) Maximum Voltage: \(V_\text{max} = 110 \mathrm{~V}\)

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Most popular questions from this chapter

The quality factor, \(Q\), of a circuit can be defined by \(Q=\omega_{0}\left(U_{E}+U_{B}\right) / P .\) Express the quality factor of a series RLC circuit in terms of its resistance \(R\), inductance \(L\), and capacitance \(C .\)

Laboratory experiments with series RLC circuits require some care, as these circuits can produce large voltages at resonance. Suppose you have a 1.00 - \(\mathrm{H}\) inductor (not difficult to obtain) and a variety of resistors and capacitors. Design a series RLC circuit that will resonate at a frequency (not an angular frequency) of \(60.0 \mathrm{~Hz}\) and will produce at resonance a magnification of the voltage across the capacitor or the inductor by a factor of 20.0 times the input voltage or the voltage across the resistor.

A series RLC circuit is in resonance when driven by a sinusoidal voltage at its resonant frequency, \(\omega_{0}=(L C)^{-1 / 2}\) But if the same circuit is driven by a square-wave voltage (which is alternately on and off for equal time intervals), it will exhibit resonance at its resonant frequency and at \(\frac{1}{3}, \frac{1}{5}\), \(\frac{1}{7}, \ldots,\) of this frequency. Explain why.

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 circuit contains a \(100 .-\Omega\) resistor, a \(0.0500-\mathrm{H}\) inductor, a \(0.400-\mu \mathrm{F}\) capacitor, and a source of time-varying emf connected in series. The time-varying emf corresponds to \(V_{\mathrm{rms}}=50.0 \mathrm{~V}\) at a frequency of \(2000 . \mathrm{Hz}\). a) Determine the current in the circuit. b) Determine the voltage drop across each component of the circuit. c) How much power is drawn from the source of emf?
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