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Chapter 19: Problem 30

An automobile starter motor is connected to a 12.0 \(\mathrm{V}\) battery. When the starter is activated it draws 150 \(\mathrm{A}\) of current, and the battery voltage drops to 7.0 \(\mathrm{V} .\) What is the battery's internal resistance?

Short Answer

Expert verified
The internal resistance is 0.033 惟.

Step by step solution

01

Understanding the Problem

We are given a 12.0 V battery which drops to 7.0 V when a 150 A current is drawn. We need to find the internal resistance of the battery.
02

Applying Ohm's Law

Ohm's law relates voltage \(V\), current \(I\), and resistance \(R\) by the formula \(V = IR\). For internal resistance \(r\), the voltage drop when the current flows through the battery is determined by \(Vr = Ir \) where \(Vr\) is the voltage drop inside the battery due to internal resistance.
03

Calculating Voltage Drop

The initial battery voltage is 12.0 V and the voltage when the motor is running is 7.0 V. The drop in voltage due to internal resistance is \( Vr = 12.0 \, \text{V} - 7.0 \, \text{V} = 5.0 \, \text{V} \).
04

Solving for Internal Resistance

Using the current \( I = 150 \, \text{A} \) and the internal voltage drop \( Vr = 5.0 \, \text{V} \), apply Ohm's law for internal resistance: \( Vr = I \cdot r \). Substitute the known values into the formula: \( 5.0 \, \text{V} = 150 \, \text{A} \cdot r \).
05

Performing the Calculation

Divide both sides of the equation by 150 A to solve for \( r \): \( r = \frac{5.0 \, \text{V}}{150 \, \text{A}} = 0.033 \, \Omega \). Thus, the internal resistance of the battery is \( 0.033 \, \Omega \).

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Key Concepts

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

Internal Resistance
Every electric device, including batteries, has a property known as internal resistance. This is a kind of resistance that exists within the device itself and can affect the performance of the electrical system. When a device powers a load, such as an automobile starter motor, current flows through this internal resistance. As a result, some of the voltage provided by the battery is "lost," resulting in a lower voltage available to the external circuit. In our exercise, the car battery had an internal resistance that caused a voltage drop from its ideal 12.0 V down to 7.0 V when the starter motor was engaged. This meant that not all of the electrical energy was used effectively in starting the engine, as some of it was used to overcome the battery's own resistance. To calculate this internal resistance, Ohm's law can be applied. The voltage drop is the difference between the original battery voltage and the observed lower voltage when the current flows through it.
Voltage Drop
Voltage drop is a term describing the reduction in voltage in an electrical circuit between the power source and the load. Whenever current flows through a component with resistance, it experiences a drop in voltage. This is because energy is used in overcoming the resistance, which manifests as a reduction in voltage. For instance, in our problem, when the starter motor of the car is activated, the voltage drop inside the battery due to its internal resistance is given by the difference between the initial voltage of 12.0 V and the final voltage of 7.0 V, resulting in a 5.0 V drop. Understanding voltage drop is crucial in real-world applications to ensure that devices receive the proper voltage for efficient operation. Large voltage drops may lead to inefficient operation or failure to power the device properly.
Electric Current
Electric current, represented by the symbol \( I \) and measured in amperes (A), is the flow of electric charge through a circuit. It's like the flow of water through pipes, where the electric current represents the amount of charge flowing per unit time. In electrical circuits, this current flows through conductive paths, such as wires, and components like resistors and batteries. In our example, when the car starter motor is activated, it draws a current of 150 A from the battery. The magnitude of this current impacts the voltage drop across the battery's internal resistance.Without adequate current, devices may not receive sufficient energy to function. Too much current, however, can lead to overheating and damage. This is why understanding how electric current interacts with resistance鈥攊ncluding internal resistance鈥攊s vital for designing and troubleshooting electrical circuits.

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Most popular questions from this chapter

In an ionic solution, a current consists of \(\mathrm{Ca}^{2+}\) ions (of charge \(+2 e )\) and \(\mathrm{Cl}^{-}\) ions (of charge \(-e )\) traveling in opposite directions. If \(5.11 \times 10^{18} \mathrm{Cl}^{-}\) ions go from \(A\) to \(B\) every 0.50 min, while \(3.24 \times 10^{18} \mathrm{Ca}^{2+}\) ions move from \(B\) to \(A\), what is the current (in mA) through this solution, and in which direction \((\) from \(A\) to \(B\) or from \(B\) to \(A)\) is it going?

A 6.00 \(\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a 4500\(\Omega\) resistor and a 500 \(\mathrm{V}\) emf source with negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor, (b) the voltage drop across the resistor, (c) the charge on the capacitor, and (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants), what are the values of the preceding four quantities?

If we model the pore of a hypothetical ion channel spanning the membrane of a nerve cell as a cylinder 0.3 \(\mathrm{nm}\) long with a radius of 0.3 \(\mathrm{nm}\) filled with a fluid of resistivity \(100 \Omega \mathrm{cm},\) what is the conductance of the channel? (Be careful with units). A. 1 \(\mathrm{G} \Omega\) B. 1 \(\mathrm{nS}\) C. 100 \(\mathrm{G\Omega}\) D. 100 \(\mathrm{nS}\)

The following measurements of current and potential difference were made on a resistor constructed of Nichrome \(^{\mathrm{TM}}\) wire, where \(V_{a b}\) is the potential difference across the wire and \(I\) is the current through it: $$\begin{array}{ccccc}{I(\mathrm{A})} & {0.50} & {1.00} & {2.00} & {4.00} \\\ {V_{a b}(\mathrm{V})} & {1.94} & {3.88} & {7.76} & {15.52}\end{array}$$ (a) Graph \(V_{a b}\) as a function of \(I .\) (b) Does Ohm's law apply to Nichromet"y? How can you tell? (c) What is the resistance of the resistor in ohms?

A 500.0\(\Omega\) resistor is connected in series with a capacitor. What must be the capacitance of the capacitor to produce a time constant of 2.00 \(\mathrm{s} ?\)
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