Problem 11 - College Physics

Question

Find the rms and peak currents in a \(40-\mathrm{W}, 120-\mathrm{V}\) lightbulb. What is its resistance?

Step-by-Step Solution

Answer

Resistance is 360 Ω; RMS current is 0.333 A; Peak current is 0.471 A.

Step 1: Understanding the Question

We are given a 40-Watt light bulb that operates at 120 Volts. We need to find the root mean square (RMS) and peak currents, as well as the bulb's resistance. RMS current is the effective current value, while peak current is the maximum current. Resistance (R) can be calculated using the formula relating power (P), voltage (V), and resistance.

Step 2: Calculate Resistance

We use the power formula for resistance: \[ R = \frac{V^2}{P} \]Plugging in the given values, \[ R = \frac{120^2}{40} = \frac{14400}{40} = 360 \; \Omega \]Thus, the resistance of the lightbulb is 360 ohms.

Step 3: Calculate RMS Current

RMS current \(I_{rms}\) can be calculated using the formula:\[ P = V \times I_{rms} \]Rearranging to find \(I_{rms}\), \[ I_{rms} = \frac{P}{V} = \frac{40}{120} = \frac{1}{3} \; \text{A} \]The RMS current is 0.333 A.

Step 4: Calculate Peak Current

Peak current \(I_{peak}\) can be found from the RMS current using the relationship:\[ I_{peak} = I_{rms} \times \sqrt{2} \]Plugging in the value for \(I_{rms}\),\[ I_{peak} = 0.333 \times \sqrt{2} \approx 0.471 \; \text{A} \]The peak current is approximately 0.471 A.

Key Concepts

RMS Current

RMS stands for Root Mean Square and is crucial in understanding AC (Alternating Current) systems. RMS Current tells us about the effective or equivalent direct current (DC) value of an AC current. This is important because it gives us a figure that can be directly compared to DC currents.
The RMS value is calculated by taking the square root of the average of the squares of the instantaneous current values over a cycle. In simpler terms, it's the value that would produce the same heating effect in a resistor as a given AC current if only DC were flowing through that resistor.
To find the RMS current, we use the power formula:

• Power (\( P \)) = Voltage (\( V \)) \( \times \) RMS Current (\( I_{rms} \))
• Rearranging gives: \( \frac{P}{V} = I_{rms} \)
• For a 40-W lightbulb at 120 V: \( I_{rms} = \frac{40}{120} = 0.333 \, A \)

Understanding RMS helps in ensuring that the electrical devices work efficiently and safely. It's a representation of AC that is comparable to DC values, allowing for proper functioning in circuits.

Peak Current

Peak Current refers to the maximum current value that occurs during a single cycle of AC. It is an important concept because it is the highest point of instantaneous electrical charge flowing through an electrical circuit. Although AC current alternates and goes through cycles of highs and lows, knowing the peak current can help anticipate the maximum load on a circuit.
In AC systems, peak current (\( I_{peak} \)) can be calculated from the RMS current using the square root of 2 as a multiplier:

• \( I_{peak} = I_{rms} \times \sqrt{2} \)
• For our example, \( I_{peak} = 0.333 \times \sqrt{2} \approx 0.471 \, A \)

Knowing the peak current prepares us for the highest possible electrical load, ensuring that circuits aren't accidentally overloaded and that all parts of the circuit are capable of handling these peak values.

Resistance Calculation

Resistance is a measure of the opposition to the flow of electric current in a circuit. In an electrical context, it helps us understand how easily or difficultly electricity can pass through materials. Understanding resistance is key to designing circuits that use electricity efficiently and safely.
The resistance (\( R \)) of a device can be calculated using the formula for electrical power:

• \( R = \frac{V^2}{P} \)
• Where \( V \) is the voltage and \( P \) is the power.
• In our lightbulb example: \( R = \frac{120^2}{40} = 360 \, \Omega \)

Understanding the resistance helps in ensuring that devices draw the correct amount of current and power, maintaining the intended brightness of a light bulb, for example, and preventing overheating or damage to electrical components.