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Chapter 26: Problem 5

What is the potential drop across an electric hot plate that draws \(5.0\) A when its hot resistance is \(24 \Omega\) ? $$ V=I R=(5.0 \mathrm{~A})(24 \Omega)=0.12 \mathrm{kV} $$

Short Answer

Expert verified
The potential drop across the hot plate is \(120 \text{ V}\).

Step by step solution

01

Identify Given Values

First, identify the values provided in the problem: we have a current \(I = 5.0 \text{ A}\) and a resistance \(R = 24 \Omega\). Our goal is to find the potential drop, denoted as \(V\), across the hot plate.
02

Use Ohm's Law

Apply Ohm's Law, which states that the potential drop \(V\) across a resistance \(R\) is given by \(V = I \cdot R\). This formula relates voltage, current, and resistance.
03

Perform the Calculation

Substitute the known values into the equation: \(V = (5.0 \text{ A}) \cdot (24 \Omega)\). Calculate the result to find the voltage.
04

Solve for V

Multiplying the current by the resistance: \(V = 5.0 \times 24 = 120 \text{ V}\). Thus, the potential drop across the electric hot plate is \(120 \text{ V}\).

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Key Concepts

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

Electric Current
Electric current is like the flow of water through a pipe. It refers to the movement of electric charge through a conductor between two points. You typically measure electric current in amperes (A). This is just a fancy word for how many electrons are moving past a point in a circuit every second. Think of electric current as the amount of electricity flowing through the wires, similar to the flow of water in a river.
  • When there's more current, it means more electricity is flowing.
  • Less current means less flow.
In our original problem, the electric hot plate has a current of 5.0 A, meaning 5 coulombs of charge are passing through the plate each second. This is important because the current helps determine how much energy is used or transformed by the device.
Resistance
Resistance is like a roadblock for electric current. It's a property of a material or object that resists the flow of electric current, making it harder for the electrons to move through. Measured in ohms (惟), resistance determines how much a given current will reduce as it passes through an object.
  • High resistance means the current has a harder time flowing.
  • Low resistance means the current flows more easily.
In our problem, the hot plate has a resistance of 24 惟. This means there is some opposition to the flow of the electric current as it moves through the coils of the hot plate. Understanding resistance is key, as it directly affects how much voltage you will need for a given current, based on the relationship described by Ohm鈥檚 Law.
Potential Drop
The potential drop is the difference in electrical potential energy between two points in a circuit. It's like the height difference between the top and bottom of a slide: the taller the slide, the more potential energy you have when you slide down. You can think of it as the 'push' needed to drive the current through a resistor. In electrical terms, the potential drop is measured in volts (V), and it's what powers devices like the hot plate in the problem.
  • The bigger the potential drop, the more energy is transferred.
  • In the problem, the potential drop across the hot plate is 120 V.
This potential drop is critical for ensuring the device operates properly by providing it with enough energy to function.
Voltage Calculation
Voltage calculation is often done using Ohm's Law, which states that the potential drop (V) across a conductor between two points is equal to the electric current (I) flowing through the conductor multiplied by its resistance (R). So, the formula is given by: \[ V = I \times R \]In this formula:
  • \(V\) is the potential drop or voltage in volts.
  • \(I\) is the current in amperes.
  • \(R\) is the resistance in ohms.
In our example, when we substitute the values into the formula: \(V = 5.0 \, \text{A} \times 24 \, \Omega = 120 \, \text{V}\). Thus, the potential drop across the hot plate comes out to be 120 V, meaning that's how much 'push' is being applied to the electric charges to make them move through the resistance of the hot plate.

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Most popular questions from this chapter

A battery charger supplies a current of \(10 \mathrm{~A}\) to charge a storage battery that has an open-circuit voltage of \(5.6 \mathrm{~V}\). If the voltmeter connected across the charger reads \(6.8 \mathrm{~V}\), what is the internal resistance of the battery at this time?

How many electrons per second pass through a section of wire carrying a current of \(0.70 \mathrm{~A}\) ?

Number 10 wire has a diameter of \(2.59 \mathrm{~mm}\). How many meters of number 10 aluminum wire are needed to give a resistance of \(1.0\) \(\Omega ? \rho\) for aluminum is \(2.8 \times 10^{-8} \Omega \cdot \mathrm{m}\). From \(R=\rho L / A\) $$ L=\frac{R A}{\rho}=\frac{(1.0 \Omega)(\pi)\left(2.59 \times 10^{-3} \mathrm{~m}\right)^{2} / 4}{2.8 \times 10^{-8} \Omega \cdot \mathrm{m}}=0.19 \mathrm{~km} $$

A direct-current generator has an emf of \(120 \mathrm{~V}\); that is, its terminal voltage is \(120 \mathrm{~V}\) when no current is flowing from it. At an output of \(20 \mathrm{~A}\), the terminal potential is \(115 \mathrm{~V}\). (a) What is the internal resistance \(r\) of the generator? \((b)\) What will be the terminal voltage at an output of \(40 \mathrm{~A}\) ? The situation is much like that shown in Fig. \(26-3 .\) Now, however, \(\varepsilon=120 \mathrm{~V}\) and \(I\) is no longer \(25 \mathrm{~A}\). (a) In this case, \(I=20 \mathrm{~A}\) and the p.d. from \(A\) to \(B\) is \(115 \mathrm{~V}\). Therefore, $$ 115 \mathrm{~V}=+120 \mathrm{~V}-(20 \mathrm{~A}) r $$ from which \(r=0.25 \Omega\). (b) Now \(I=40\) A. So Terminal p.d. \(=\varepsilon-I r=120 \mathrm{~V}-(40 \mathrm{~A})(0.25 \Omega)=110 \mathrm{~V}\)

Compute the resistance of \(180 \mathrm{~m}\) of silver wire having a cross section of \(0.30 \mathrm{~mm}^{2}\). The resistivity of silver is \(1.6 \times 10^{-8} \Omega \cdot \mathrm{m}\).
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