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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