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A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance of each bulb does not vary with current. (\(Note:\) This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light bulb dissipates 25 W when connected to a 120-V line.) (a) Find the current through the bulbs. (b) Find the power dissipated in each bulb. (c) One bulb burns out very quickly. Which one? Why?

Short Answer

Expert verified
(a) The current through the bulbs is approximately 0.769 A. (b) Power dissipated: 60-W bulb = 142 W, 200-W bulb = 42.5 W. (c) The 60-W bulb burns out quickly due to excess power dissipation.

Step by step solution

01

Calculate Resistance of Each Bulb

First, calculate the resistance of each bulb using the power formula. The resistance, \( R \), can be found using the formula: \[ R = \frac{V^2}{P} \]For the 60-W bulb: \[ R_1 = \frac{120^2}{60} \approx 240 \ \Omega \]For the 200-W bulb: \[ R_2 = \frac{120^2}{200} = 72 \ \Omega \]
02

Calculate Total Resistance in Series

In a series circuit, the total resistance \( R_{total} \) is the sum of the individual resistances: \[ R_{total} = R_1 + R_2 \]\[ R_{total} = 240 \ \Omega + 72 \ \Omega = 312 \ \Omega \]
03

Calculate Current Through the Circuit

Using Ohm's Law, the current \( I \) can be calculated as: \[ I = \frac{V}{R_{total}} \]The total voltage \( V \) is 240 V:\[ I = \frac{240}{312} \approx 0.769 \ A \]
04

Calculate Power Dissipated by Each Bulb

Power dissipated by a resistor in a circuit is given by:\[ P = I^2R \]For the 60-W bulb (240 Ω):\[ P_1 = (0.769)^2 \times 240 \approx 142 \ W \]For the 200-W bulb (72 Ω):\[ P_2 = (0.769)^2 \times 72 \approx 42.5 \ W \]
05

Determine Which Bulb Burns Out Quickly

The 60-W bulb burns out quickly because it dissipates significantly more power (142 W) than its rated power (60 W), leading to overheating.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistance Calculation
When working with electric circuits, understanding how to calculate resistance is essential. Resistance is a measure of how much an object opposes the flow of current through it. In the provided exercise, to find the resistance of each bulb, we use the formula:
\[ R = \frac{V^2}{P} \]
  • For the 60-W, 120-V bulb, the resistance \( R_1 \) is calculated as \( R_1 = \frac{120^2}{60} = 240 \ \Omega \).
  • For the 200-W, 120-V bulb, the resistance \( R_2 \) is calculated as \( R_2 = \frac{120^2}{200} = 72 \ \Omega \).
Calculating resistance like this allows us to understand how each component in a circuit will behave when connected to power.
If you picture water flowing through pipes, resistance is like the narrowing or widening of a pipe, impacting how easily water can flow through.
Ohm's Law
Ohm's Law is a crucial principle in electronics. It relates voltage, current, and resistance in an electric circuit. The formula for Ohm's Law is:
\[ V = IR \]This tells us that the voltage (\( V \)) is the product of the current (\( I \)) and the resistance (\( R \)). In the given exercise, you can rearrange the formula to find current:
\[ I = \frac{V}{R_{total}} \] Where \( R_{total} \) is the sum of all resistances in the series circuit as calculated. For a 240 V line and a total resistance of 312 \( \Omega \), we find:
\[ I = \frac{240}{312} \approx 0.769 \ A \]Understanding Ohm's Law helps in predicting how changes in one quantity, like resistance, affect another, like current. It's foundational for analyzing any electric circuit.
Power Dissipation
Power dissipation is the process through which electrical energy is converted to heat energy in a circuit. This is crucial because excess power can overheat and damage components. The power dissipated through a resistor is calculated using:
\[ P = I^2 R \]In the exercise, we calculated the power dissipation for each bulb:
  • For the 60-W bulb with 240 \( \Omega \): \[ P_1 = (0.769)^2 \times 240 \approx 142 \ W \]
  • For the 200-W bulb with 72 \( \Omega \): \[ P_2 = (0.769)^2 \times 72 \approx 42.5 \ W \]
The 60-W bulb dissipates more power than its design can handle, leading it to burn out faster. Calculating power dissipation ensures that circuit components operate within safe limits and helps prevent damage.
Series Circuit Analysis
Series circuits are characterized by components connected end-to-end, so current flows through each component consecutively. A key feature of series circuits is that the total resistance is the sum of individual resistances:
\[ R_{total} = R_1 + R_2 + \ldots + R_n \]In our exercise:
  • \( R_{total} = 240 \ \Omega + 72 \ \Omega = 312 \ \Omega \)
Another critical aspect is that the same current flows through all components. Understanding series circuits helps analyze how electric current is distributed among connected devices and allows us to predict electrical behaviour under different conditions. Series circuit analysis is fundamental for designing and troubleshooting electrical systems.

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Most popular questions from this chapter

A galvanometer having a resistance of 25.0 \(\Omega\) has a 1.00-\(\Omega\) shunt resistance installed to convert it to an ammeter. It is then used to measure the current in a circuit consisting of a 15.0-\(\Omega\) resistor connected across the terminals of a 25.0-V battery having no appreciable internal resistance. (a) What current does the ammeter measure? (b) What should be the \(true\) current in the circuit (that is, the current without the ammeter present)? (c) By what percentage is the ammeter reading in error from the \(true\) current?

An emf source with \(\varepsilon =\) 120 V, a resistor with \(R =\) 80.0 \(\Omega\), and a capacitor with \(C =\) 4.00 \(\mu\)F are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

A 4.60-\(\mu\)F capacitor that is initially uncharged is connected in series with a 7.50-k\(\Omega\) resistor and an emf source with \(\varepsilon =\) 245 V and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)-(d)?

A resistor with \(R_1 =\) 25.0 \(\Omega\) is connected to a battery that has negligible internal resistance and electrical energy is dissipated by \(R_1\) at a rate of 36.0 W. If a second resistor with \(R_2 =\) 15.0 \(\Omega\) is connected in series with \(R_1\), what is the total rate at which electrical energy is dissipated by the two resistors?

A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 M\(\Omega\). After a time of 4.00 s the voltmeter reads 3.0 V. What are (a) the capacitance and (b) the time constant of the circuit?

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