/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 A watch uses energy at the rate ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A watch uses energy at the rate of \(240 \mu \mathrm{W}\). What current does it draw from its 1.5 -V battery?

Short Answer

Expert verified
The current the watch draws from its 1.5 V battery is \(160 µA\).

Step by step solution

01

Understand the given variables

The power of the watch (P) is given as \(240 \mu W\) which is equivalent to \(0.00024 W\). The voltage of the battery (V) is given as 1.5 V. The aim is to find the current (I).
02

Use the formula P=IV

From Ohm’s Law, Power (P) is expressed as P=IV, where I is the current and V is the voltage. We need to solve this formula for current (I), given that we know the power (P) and the voltage (V). The rearranged equation is I=P/V.
03

Substitute the values

Substitute the given power and voltage into the formula. So \(I=0.00024/1.5\). As a result, the current I that the watch draws from its battery is \(0.00016 A\). It's more common to express it in microamperes (µA), which is \(0.00016 A * 10^6 = 160 µA\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Electric Current
Electric current is the flow of electric charge. In most cases, the charge moves through a conductor like a wire. Electric current is measured in amperes (A), which indicates how much charge flows through a given point in a certain amount of time. When we talk about a watch drawing current, we refer to how much electric charge it is using to operate.
The current in a circuit can be direct (DC) or alternating (AC), but in the case of batteries like the one in the watch, it's usually DC. Current direction is often shown as flowing from the positive to the negative terminal, even though electrons move in the opposite direction.
Understanding electric current helps us analyze how electrical devices use power, predicting their energy consumption and efficiency.
Ohm's Law
Ohm's Law is a fundamental principle in the field of electronics and electrical engineering. It sets out the relationship between voltage, current, and resistance in a circuit, expressed in the formula: \[ V = IR \]Where:
  • \( V \) is the voltage (in volts),
  • \( I \) is the current (in amperes), and
  • \( R \) is the resistance (in ohms).
By using Ohm's Law, one can easily calculate any of the three variables if the other two are known. This was essential in our example, along with the power formula, to find the current drawn by the watch. Knowing the power and voltage, rearranging the power formula \( P = IV \) allowed us to solve for current (\( I \)) using \( I = \frac{P}{V} \). This highlights the versatility and importance of Ohm's Law in solving practical physics problems.
Electrical Power
Electrical power measures the rate at which electrical energy is consumed by an appliance or device. It is given in watts (W), which indicates how fast energy is being used or supplied in a circuit. Power is a key factor in determining what devices can run on a particular battery or power supply.
In our exercise, the watch uses power at a rate of \(240 \mu W\). To express this in watts, we converted it as \(0.00024 W\). To calculate the current, knowing the voltage and power, is a demonstration of how electrical power calculations are applied in real-world scenarios.
The relationship between power, voltage, and current is expressed by the formula: \[ P = IV \],which emphasizes how much work is done or how much energy is consumed in a given time.
Voltage
Voltage, or electrical potential difference, is the electrical force that drives an electric current between two points. It's measured in volts (V). Essentially, voltage is the push that causes the charges to move in a wire.
In the example given, the watch is powered by a 1.5-V battery. This voltage is what enables the current to flow through the watch, providing the energy required for it to function. The higher the voltage, the more energy each charge carries.
Understanding voltage is crucial as it impacts the amount of power a device can use. Devices and batteries are designed with specific voltage requirements to ensure optimal performance and safety.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Passage Problems A brownout occurs when an electric utility can't supply enough power to meet demand. Rather than cut off some customers completely, the utility reduces the voltage across its system. Brownouts are most likely on hot summer days, when heavy air-conditioning loads drive up demand for electricity. In a particular brownout, the utility reduces the voltage by \(10 \%.\) During the brownout, the current in conductors whose resistance is nearly independent of temperature a. decreases by approximately \(10 \%.\) b. decreases by approximately \(20 \%.\) c. decreases by approximately \(5 \%.\) d. You can't tell without knowing the resistance.

How must the diameters of copper and aluminum wire be related if they're to have the same resistance per unit length?

Magnetic effects involving Jupiter's moon Io result in an effective voltage of about \(400 \mathrm{kV},\) which drives current of some \(5 \mathrm{MA}\) between Io and Jupiter's polar regions. At Jupiter the current produces auroras analogous to those on Earth, as well as powerful bursts of radio waves that help radio astronomers analyze the Jovian current system. Estimate the total power associated with this current system, and compare with the 2 -TW rate at which humankind consumes electrical energy.

You're working on a new high-speed rail system. It uses 6000 -horsepower electric locomotives, getting power from a single overhead wire with resistance \(15 \mathrm{m} \Omega / \mathrm{km},\) at \(25 \mathrm{kV}\) potential relative to the track. Current returns through the track, whose resistance is negligible. Energy-efficiency standards call for no more than \(3 \%\) power loss in the wire. How far from the power plant can the train go and still meet this standard?

You're estimating costs for a new power line with your company's financial group. Engineering specifies a resistance per unit length of \(50 \mathrm{m} \Omega / \mathrm{km} .\) The costs of copper and aluminum wire are \(\$ 4.65 / \mathrm{kg}\) and \(\$ 2.30 / \mathrm{kg}\) and their densities are \(8.9 \mathrm{g} / \mathrm{cm}^{3}\) and \(2.7 \mathrm{g} / \mathrm{cm}^{3},\) respectively. Which material is more economical?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.