/*! 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 22 An automobile starter motor has ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An automobile starter motor has an equivalent resistance of \(0.0500 \Omega\) and is supplied by a 12.0 -V battery with a \(0.0100-\Omega\) internal resistance. (a) What is the current to the motor? (b) voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add \(0.0900 \Omega\) to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)

Short Answer

Expert verified
In short, (a) the current to the motor under normal conditions is \(200.0 A\), (b) the voltage applied to the motor is \(10.0 V\), and (c) the power supplied to the motor is \(2.00 \times10^{3} W\). When the battery connections are corroded, (d) the current to the motor is reduced to \(100.0 A\), the voltage applied to the motor is \(5.00 V\), and the power supplied to the motor drops to \(5.00 \times 10^{2} W\).

Step by step solution

01

Calculate the total resistance in the circuit (normal conditions)

The total resistance in the circuit is the sum of the starter motor's equivalent resistance and the battery's internal resistance. \(R_{total} = R_{motor} + R_{battery}\) \(R_{total} = 0.0500 \Omega + 0.0100 \Omega\)
02

Calculate the current flowing to the motor (normal conditions)

Using Ohm's Law (Voltage = Current × Resistance), we can calculate the current flowing to the motor. \(I = \frac{V}{R_{total}}\) \(I = \frac{12.0 V}{0.0600 \Omega}\)
03

Calculate the voltage applied to the motor (normal conditions)

The voltage applied to the motor is the voltage drop across the motor's resistance. This can be calculated using Ohm's Law. \(V_{motor} = I × R_{motor}\)
04

Calculate the power supplied to the motor (normal conditions)

The power supplied to the motor can be calculated using the power equation (Power = Voltage × Current). \(P_{motor} = V_{motor} × I\)
05

Calculate the total resistance when the battery connections are corroded

When the battery connections are corroded, the circuit has an additional resistance of 0.0900 Ω. Taking that into account, the total resistance in the circuit is as follows: \(R_{total \; corroded} = R_{motor} + R_{battery} + R_{corrosion}\) \(R_{total \; corroded} = 0.0500 \Omega + 0.0100 \Omega + 0.0900 \Omega\)
06

Calculate the current flowing to the motor when the battery connections are corroded

Using Ohm's Law with the total corroded resistance, we can calculate the current flowing to the motor. \(I_{corroded} = \frac{V}{R_{total \; corroded}}\)
07

Calculate the voltage applied to the motor when the battery connections are corroded

The voltage applied to the motor in the corroded case is the voltage drop across the motor's resistance. This can be calculated using Ohm's Law. \(V_{motor \; corroded} = I_{corroded} × R_{motor}\)
08

Calculate the power supplied to the motor when the battery connections are corroded

The power supplied to the motor in the corroded case can be calculated using the power equation (Power = Voltage × Current). \(P_{motor \; corroded} = V_{motor \; corroded} × I_{corroded}\)

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.

Electrical Resistance
Electrical resistance is a fundamental concept in the study of electronics and physics. It measures the degree to which a substance opposes the flow of electric current. Think of it like a water pipe: if the pipe is narrow, it's harder for water to flow through—the same goes for electricity in a conductor. The resistance is dependent on the material, length, and cross-sectional area of the conductor.

The standard unit for measuring resistance is the ohm (Ωg)). In the context of the automobile starter motor from our exercise, the equivalent resistance is quite low (0.0500 Ω), meaning that it allows a substantial amount of current to flow with minimal opposition. However, when the connections are corroded, resistance increases (0.0900 Ω added), mimicking a narrower pipe and thus reducing the current.
Electric Current
Electric current is the flow of electric charge. In a practical sense, it's the electricity moving through a wire to power devices like the starter motor in the exercise. The current is measured in amperes (amps, A) and can be visualized as the amount of water flowing through a pipe—the greater the flow, the higher the current. Using Ohm's Law (V = IR), we can calculate the current by dividing the voltage by the total resistance. On normal conditions with low resistance, a strong current of 200 A is sent to the motor, capable of starting the car effectively. However, with added resistance due to corrosion, this current drops significantly—which is undesirable in car starter systems.
Electrical Power
Electrical power is essentially the rate at which electrical energy is transferred by an electric circuit. The unit of power is the watt (W). In our car example, the power supplied to the starter motor affects how quickly and with what force the motor turns over the engine. The power equation (Power = Voltage × Current), shows that both voltage and current affect the total power. Under normal conditions, the power supplied to the starter motor is substantial, thereby ensuring that the motor has enough force to start the engine. When resistance is added due to corrosion, the current reduces, ultimately decreasing the power supplied to the motor. Less power means the starter motor is less capable of turning over the engine, illustrating the importance of maintaining a low-resistance path in automotive electrical systems.
Voltage Drop
Voltage drop is the loss of voltage as the electrical current moves through a conductor. It's similar to losing water pressure in a long hose due to friction. In electrical terms, voltage drop can signal energy being converted to another form, like heat, due to the resistance of the material. Using Ohm's Law, a voltage drop can be calculated by multiplying the current by the resistance (V = IR). In our example, the voltage drop tells us how much voltage is effectively being applied to the starter motor itself. The starter motor initially receives almost the full 12.0 V from the battery, but as resistance from corrosion builds up, a larger portion of the voltage is lost before reaching the motor, resulting in reduced efficiency and performance.

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

A voltage occurs across an open switch. What is the power dissipated by the open switch?

Analog meters use a galvanometer, which essentially consists of a coil of wire with a small resistance and a pointer with a scale attached. When current runs through the coil, the pointer turns; the amount the pointer turns is proportional to the amount of current running through the coil. Galvanometers can be used to make an ammeter if a resistor is placed in parallel with the galvanometer. Consider a galvanometer that has a resistance of \(25.00 \Omega\) and gives a full scale reading when a \(50-\mu\) A current runs through it. The galvanometer is to be used to make an ammeter that has a full scale reading of \(10.00 \mathrm{A},\) as shown below. Recall that an ammeter is connected in series with the circuit of interest, so all \(10 \mathrm{A}\) must run through the meter. (a) What is the current through the parallel resistor in the meter? (b) What is the voltage across the parallel resistor? (c) What is the resistance of the parallel resistor?

What are the advantages and disadvantages of connecting batteries in series? In parallel?

An 1800-W toaster, a 1400-W speaker, and a 75-W lamp are plugged into the same outlet in a 15 -A fuse and 120-V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current is drawn by each device? (b) Will this combination blow the \(15-\mathrm{A}\) fuse?

Do batteries in a circuit always supply power to a circuit, or can they absorb power in a circuit? Give an example.

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.