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An unknown resistor is connected between the terminals of a \(3.00 \mathrm{~V}\) battery. Energy is dissipated in the resistor at the rate of \(0.540 \mathrm{~W}\). The same resistor is then connected between the terminals of a \(1.50 \mathrm{~V}\) battery. At what rate is energy now dissipated?

Short Answer

Expert verified
The energy is now dissipated at a rate of 0.135 W.

Step by step solution

01

Understand the Formula for Power Dissipation

The power dissipated by a resistor is given by the formula \( P = \frac{V^2}{R} \), where \( P \) is the power in watts, \( V \) is the voltage in volts, and \( R \) is the resistance in ohms. This means the power is proportional to the square of the voltage.
02

Calculate Resistance with Initial Conditions

Initially, a 3.00 V battery dissipates energy at 0.540 W. Using the formula \( P = \frac{V^2}{R} \), calculate \( R \):\[R = \frac{V^2}{P} = \frac{(3.00)^2}{0.540}\]Solving this gives \( R \approx 16.67 \) ohms.
03

Calculate New Power Dissipation

Now, using a 1.50 V battery with the same resistor, we use the power formula again:\[P = \frac{V^2}{R} = \frac{(1.50)^2}{16.67}\]Calculate this to find the new power dissipation.
04

Perform the Final Calculation

Substitute the values into the equation:\[P = \frac{2.25}{16.67} \approx 0.135 \text{ W}\]Thus, when the resistor is connected to a 1.50 V battery, the power dissipation is 0.135 W.

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Key Concepts

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

Ohm's Law
Understanding Ohm's Law is crucial for grasping how electrical currents flow through circuits. This law is a fundamental principle in the study of electronics. Simply put, Ohm's Law states that the current (\(I\)) running through a conductor between two points is directly proportional to the voltage (\(V\)) across the two points and inversely proportional to the resistance (\(R\)) of the circuit. This can be summarized in the equation:
  • \(V = IR\)
When you're working with circuits, understanding this relation is vital. This formula allows you to predict how changes in voltage or resistance will impact the current. In exercises involving power dissipation, knowing how voltage and resistance interrelate through Ohm's Law helps decode how electrical energy converts into heat or motion.As such, if you know any two of the three variables (\(V\), \(I\), \(R\)), you can effortlessly calculate the third. This equation can also rearrange to solve for current or resistance, depending on the known variables, helping solve a wide array of practical problems in physics and engineering.
Electrical Resistance
Resistance is a key concept when exploring electrical circuits. It refers to how much a material or device opposes the flow of electric current. The unit of resistance is the Ohm (\(\Omega\)), and it plays a vital role in determining the behavior of circuits.When a resistor, like the one in the example problem, is connected in a circuit, it limits the amount of current that can pass through. The power dissipation in a resistor, as might occur when using a battery, is typically undesirable, as it converts electric energy into heat that doesn't accomplish useful work. The formula for calculating power dissipation in a resistance helps highlight how resistance impacts energy loss:
  • \(P = \frac{V^2}{R}\)
Here, you see that power (\(P\)) is inversely related to resistance (\(R\)). This means that as resistance increases, less power is dissipated, assuming a constant voltage. This relationship underlines why precision in calculating resistance is paramount in designing efficient circuits.
Voltage
Voltage, sometimes referred to as electric potential difference, is a measure of how much potential energy exists to move an electric current from one point to another. In the context of circuitry, it truly represents the force that pushes electrons through a conductor. Voltage is measured in volts and symbolized as \(V\).In the exercise discussed, understanding voltage's role is essential, as it directly influences how much power a resistor dissipates. When comparing a 3.00 V battery to a 1.50 V battery while keeping resistance constant, we observe a key principle: power depends on the square of the voltage. The implications of this are significant: when voltage changes, the power dissipation by the resistor alters exponentially. This is depicted in the following formula:
  • \(P = \frac{V^2}{R}\)
Thus, if voltage halves, power does not simply halve – it reduces by a factor of four, as observed in the example problem. Hence, Voltage is crucial in determining the efficacy and efficiency of electronic circuits in powering devices and ensuring safety.

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Most popular questions from this chapter

A cylindrical resistor of radius \(5.0 \mathrm{~mm}\) and length \(2.0 \mathrm{~cm}\) is made of material that has a resistivity of \(3.5 \times 10^{-5} \Omega \cdot \mathrm{m} .\) What are (a) the magnitude of the current density and (b) the potential difference when the energy dissipation rate in the resistor is \(1.0 \mathrm{~W} ?\)

A beam contains \(2.0 \times 10^{8}\) doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of \(1.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). What are the (a) magnitude and (b) direction of the current density \(\vec{J} ?\) (c) What additional quantity do you need to calculate the total current \(i\) in this ion beam?

A potential difference of \(1.20 \mathrm{~V}\) will be applied to a \(33.0 \mathrm{~m}\) length of 18 -gauge copper wire (diameter \(=0.0400\) in.). Calculate (a) the current, (b) the magnitude of the current density, (c) the magnitude of the electric field within the wire, and (d) the rate at which thermal energy will appear in the wire.

A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to \(440 \mathrm{~A} / \mathrm{cm}^{2} .\) What diameter of cylindrical wire should be used to make a fuse that will limit the current to \(0.50 \mathrm{~A}\) ?

Thermal energy is produced in a resistor at a rate of \(100 \mathrm{~W}\) when the current is \(3.00 \mathrm{~A}\). What is the resistance?

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