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Chapter 25: Problem 39

Light Bulbs. The power rating of a light bulb (such as a \(100-\mathrm{W}\) bulb) is the power it dissipates when connected across a \(120-\mathrm{V}\) potential difference. What is the resistance of (a) a \(100-\mathrm{W}\) bulb and (b) a \(60-\mathrm{W}\) bulb? (c) How much current does each bulb draw in normal use?

Short Answer

Expert verified
100-W bulb: 144 惟, draws 0.833 A; 60-W bulb: 240 惟, draws 0.5 A.

Step by step solution

01

Understanding the Formula for Power

The power rating of a light bulb, which is given in watts (W), is calculated using the formula: \[ P = V \times I \] where \( P \) is power, \( V \) is voltage, and \( I \) is current. Rearrange this formula to solve for current: \[ I = \frac{P}{V} \]
02

Applying Ohm's Law to Find Resistance

Ohm's law states that \( V = I \times R \), where \( R \) is resistance. We can rearrange this to find resistance: \[ R = \frac{V}{I} \]. We'll first calculate the current for each bulb using the formula derived in Step 1, and then find the resistance.
03

Calculating the Resistance of the 100-W Bulb

For the \(100\)-W bulb, we use the formula for current: \[ I = \frac{100\,\text{W}}{120\,\text{V}} = 0.833\,\text{A} \]. Then, use this current to calculate resistance: \[ R = \frac{120\,\text{V}}{0.833\,\text{A}} \approx 144\,\Omega \].
04

Calculating the Resistance of the 60-W Bulb

For the \(60\)-W bulb, calculate the current: \[ I = \frac{60\,\text{W}}{120\,\text{V}} = 0.5\,\text{A} \]. Now, calculate the resistance: \[ R = \frac{120\,\text{V}}{0.5\,\text{A}} = 240\,\Omega \].
05

Determining the Current For Each Bulb

The currents have already been calculated in Steps 3 and 4: - The \(100\)-W bulb draws \(0.833\,\text{A}\).- The \(60\)-W bulb draws \(0.5\,\text{A}\).

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Key Concepts

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

Power Calculation
Power calculation in electrical circuits is an essential concept. Power, denoted by the symbol \(P\), is the rate at which electrical energy is transferred by an electric circuit. It is measured in watts (W).
  • To calculate power, we use the formula \(P = V \times I\), where \(V\) stands for voltage in volts, and \(I\) stands for current in amperes (A).
  • Here, voltage is the potential difference across the element, and the current is the flow of electric charges through it.
By rearranging the formula, we can also determine the current using the formula \(I = \frac{P}{V}\), which is particularly useful when we know the voltage and power, but need to find the current through the circuit. Breaking down this formula can greatly help in designing circuits and calculating the power requirements for various electrical devices.
Ohm's Law
Ohm's Law is one of the fundamental principles in the field of electronics. It establishes a relationship between voltage, current, and resistance in an electrical circuit. This law is described by the equation \(V = I \times R\).
  • \(V\) is the voltage across the resistance.
  • \(I\) is the current flowing through the resistance.
  • \(R\) is the resistance, which opposes the flow of current.
By manipulating Ohm's Law, we can solve for any of the three variables if the other two are known. For instance, to find the resistance \(R\), we can rearrange the equation as \(R = \frac{V}{I}\). Understanding Ohm's Law allows us to handle and manage electrical circuits effectively, predicting how a circuit will behave under different conditions.
Electrical Resistance
Electrical resistance is a concept that describes how difficult it is for electric current to flow through a conductor. It is measured in ohms (惟) and is determined by the material, length, and cross-sectional area of the conductor.
  • The symbol for resistance is \(R\).
  • In our light bulb example, resistance affects how much current flows for a given voltage.
  • The higher the resistance, the less current flows at a given voltage.

Resistance can be calculated using Ohm's Law, \(R = \frac{V}{I}\). For example, for the 100-W bulb, given the voltage and current, we calculated the resistance to be approximately 144 惟. Similarly, for the 60-W bulb, the resistance was calculated to be 240 惟. Understanding resistance helps in designing circuits with optimal performance while preserving the integrity of electronic components.
Current Calculation
Current calculation is a vital part of analyzing and understanding electrical circuits. Current, denoted as \(I\), is the flow of electric charges through a conductor, and it is measured in amperes (A).
  • The formula for calculating current is \(I = \frac{P}{V}\), where \(P\) is the power in watts, and \(V\) is the voltage in volts.
  • In the given exercise, current was calculated for each bulb.
  • For the 100-W bulb with a voltage of 120 V, the current is 0.833 A.
  • For the 60-W bulb, it is 0.5 A, calculated using the same formula.

By calculating the current, we can evaluate how much electricity flows through a component of the circuit, which is crucial for ensuring that the components operate safely within their limits, avoiding overheating and ensuring energy efficiency. This concept, along with power and resistance, is key for effective circuit management.

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Most popular questions from this chapter

Electric Eels. Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500 \(\mathrm{V}\) and produce currents of 80 \(\mathrm{mA}\) (or even larger). A typical pulse lasts for 10 \(\mathrm{ms}\) . What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

A 5.00 -A current runs through a 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm}\) ) and through a light bulb. Copper has \(8.5 \times\) \(10^{28}\) free electrons per cubic meter, (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

A lightning bolt strikes one end of a steel lightning rod, producing a \(15,000-\) A current burst that lasts for 65 \mus. The rod is 2.0 \(\mathrm{m}\) long and 1.8 \(\mathrm{cm}\) in diameter, and its other end is connected to the ground by 35 \(\mathrm{m}\) of \(8.0-\mathrm{mm}\) -diameter copper wire. (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

A source with emf \(\mathcal{E}\) and internal resistance \(r\) is connected to an external circuit. (a) Show that the power output of the source is maximum when the current in the circuit is one-half the short-circuit current of the source. (b) If the external circuit consists of a resistance \(R,\) show that the power output is maximum when \(R=r\) and that the maximum power is \(\mathcal{E}^{2} / 4 r_{1}\)

A cylindrical copper cable 1.50 \(\mathrm{km}\) long is connected across a 220.0 -V potential difference. (a) What should be its diameter so that it produces heat at a rate of 75.0 \(\mathrm{W}^{\prime}\) (b) What is the electric field inside the cable under these conditions?
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