Question

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

(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

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: $$R_1=\frac{{120}^2}{60}\approx 240\mathrm{\Omega }$$For the 200-W bulb: $$R_2=\frac{{120}^2}{200}=72\mathrm{\Omega }$$

Calculate Total Resistance in Series

In a series circuit, the total resistance $R_{total}$ is the sum of the individual resistances: $$R_{total}=R_1+R_2$$$$R_{total}=240\mathrm{\Omega }+72\mathrm{\Omega }=312\mathrm{\Omega }$$

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:$$I=\frac{240}{312}\approx 0.769A$$

Calculate Power Dissipated by Each Bulb

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

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.

Key concepts

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 $R_1=\frac{{120}^2}{60}=240\mathrm{\Omega }$.
• For the 200-W, 120-V bulb, the resistance $R_2$ is calculated as $R_2=\frac{{120}^2}{200}=72\mathrm{\Omega }$.

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 $\mathrm{\Omega }$, we find:
$$I=\frac{240}{312}\approx 0.769A$$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 $\mathrm{\Omega }$: $$P_1=(0.769{)}^2\times 240\approx 142W$$
• For the 200-W bulb with 72 $\mathrm{\Omega }$: $$P_2=(0.769{)}^2\times 72\approx 42.5W$$

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:
$$R_{total}=R_1+R_2+\dots +R_n$$In our exercise:

• $R_{total}=240\mathrm{\Omega }+72\mathrm{\Omega }=312\mathrm{\Omega }$

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.