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Chapter 9: Problem 17

Resistors are commonly rated at \(\frac{1}{8} \mathrm{W}, \frac{1}{4} \mathrm{W}, \frac{1}{2} \mathrm{W}\) 1 W and 2 W for use in electrical circuits. If a current of \(I=2.00 \mathrm{A}\) is accidentally passed through a \(R=1.00 \Omega\) resistor rated at \(1 \mathrm{W}\), what would be the most probable outcome? Is there anything that can be done to prevent such an accident?

Short Answer

Expert verified
The most probable outcome when a current of 2 A is accidentally passed through a 1 惟 resistor rated at 1 W is that the resistor will overheat and get damaged or even burn out, as the calculated power dissipated by the resistor (4 W) is greater than its rated power (1 W). To prevent such accidents, one can choose a resistor with an appropriate power rating, use a fuse or circuit breaker, or employ current-limiting protection devices in the circuit.

Step by step solution

01

Calculate the actual power dissipated by the resistor

To calculate the actual power dissipated by the resistor, we will use the formula P = I^2 * R. Given the current I = 2 A and the resistance R = 1 惟, we can calculate the power as follows: \(P = (2)^2 * (1)\) \(P = 4 \, \mathrm{W}\)
02

Compare the calculated power with the rated power

Now that we have calculated the power dissipated by the resistor (4 W), we can compare it to the rated power of the resistor, which is 1 W. Since the calculated power (4 W) is greater than the rated power (1 W), the most probable outcome is that the resistor will overheat and get damaged or even burn out.
03

Suggest some methods to prevent such accidents

There are several ways to prevent such accidents, including: 1. Choosing the right resistor: The first step is to choose a resistor with the appropriate power rating for the circuit. In this case, if we expect a current of 2 A to pass through a 1 惟 resistor, it would be better to use a resistor with a higher power rating, such as 2 W or 4 W. 2. Fuse or circuit breaker: Incorporating a fuse or a circuit breaker in the circuit can help protect the components such as resistors from damage by cutting off the current when it exceeds a specified value. 3. Current limiting protection: Another method is to use current-limiting devices (such as a current-limiting resistor or a current-limiting voltage regulator) that limit the current passing through the circuit, thereby protecting the resistor. In general, good circuit design and component selection are crucial for preventing accidents and ensuring the safety and reliability of electrical circuits.

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in the field of electronics that relates the current flowing through a conductor to the voltage across it and its resistance. It is succinctly expressed with the equation \( V = I \times R \), where \( V \) is the voltage in volts, \( I \) is the current in amperes, and \( R \) is the resistance in ohms (惟). This law is essential for designing and analyzing electrical circuits, as it helps predict how much current will flow for a given voltage and resistance. Understanding Ohm's Law is crucial for selecting the appropriate components, such as resistors, to ensure they operate within their specified ratings to prevent damage.
Electrical Circuits
Electrical circuits are paths through which electric current flows, commonly including components like resistors, capacitors, and inductors. A typical circuit might involve a power source, such as a battery, which provides the potential difference needed for current to flow. Circuits can be of two types: series and parallel, each having distinct characteristics regarding voltage and current distribution among components. Proper understanding of how circuits work is imperative to prevent overloading and ensure the longevity of the components used in the circuit, thereby avoiding accidents. When constructing or analyzing a circuit, one must always consider the maximum power that each component can safely handle.
Power Dissipation in Resistors
The power dissipation in resistors is a measure of how much electrical energy is converted into heat as the current passes through the resistor. The power \( P \) dissipated by a resistor is calculated using the formula \( P = I^2 \times R \) or \( P = \frac{V^2}{R} \) depending on whether the current \( I \) or voltage \( V \) is known. It's essential to ensure that a resistor's power rating is higher than the actual power dissipated during operation. Exceeding the power rating will lead to excessive heat, which could damage the resistor and potentially cause a failure in the circuit. Understanding this concept helps in selecting resistors with appropriate power ratings for reliable and safe circuit design.
Circuit Protection
Circuit protection is a key aspect of electrical circuit design, aimed at preventing damage to components due to overcurrent or other hazardous conditions. Protection mechanisms like fuses and circuit breakers are designed to break the circuit if the current exceeds safe levels. Current limiting components such as resistors or voltage regulators can also be used to reduce the current flow to safer levels. Ensuring adequate circuit protection not only preserves the components and prevents potential fires but also extends the longevity of the electrical system. Regularly examining and maintaining these protective devices is vital for ensuring ongoing circuit protection.

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Most popular questions from this chapter

A 20.00-ohm, 5.00-watt resistor is placed in series with a power supply. (a) What is the maximum voltage that can be applied to the resistor without harming the resistor? (b) What would be the current through the resistor?

A 0.50-W, 220- 2 resistor carries the maximum current possible without damaging the resistor. If the current were reduced to half the value, what would be the power consumed?

When \(100.0 \mathrm{V}\) is applied across a 5 -gauge (diameter 4.621 \(\mathrm{mm}\) ) wire that is \(10 \mathrm{m}\) long, the magnitude of the current density is \(2.0 \times 10^{8} \mathrm{A} / \mathrm{m}^{2}\). What is the resistivity of the wire?

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A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of \(3.00 \mathrm{A}\) and a voltage of \(110 \mathrm{V},\) a lamp that contains a \(100-\mathrm{W}\) bulb, an overhead light with a \(60-\mathrm{W}\) bulb, and various other small devices adding up to \(3.00 \mathrm{W}\). (a) Assuming the power plant that supplies \(110 \mathrm{V}\) electricity to the dorm is \(10 \mathrm{km}\) away and the two aluminum transmission cables use 0-gauge wire with a diameter of 8.252 mm, estimate the percentage of the total power supplied by the power company that is lost in the transmission. (b) What would be the result is the power company delivered the electric power at \(110 \mathrm{kV} ?\)
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