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Chapter 26: Problem 30

A dry cell delivering \(2 \mathrm{~A}\) has a terminal voltage of \(1.41 \mathrm{~V} .\) What is the internal resistance of the cell if its open-circuit voltage is \(1.59 \mathrm{~V} ?\)

Short Answer

Expert verified
The internal resistance of the cell is \(0.09 \, \Omega\).

Step by step solution

01

Understanding the Problem

We need to find the internal resistance of a dry cell. We are given the terminal voltage, the current supplied, and the open-circuit voltage of the cell.
02

Applying Ohm's Law

According to Ohm's law, the voltage across a component is given by the product of current and resistance: \( V = I imes R \). Here, we need to consider the internal characteristics of the cell.
03

Identify Relevant Formulas

The internal resistance \( r \) can be calculated using the formula \( V = \, E - I imes r \), where \( V \) is the terminal voltage, \( E \) is the open-circuit voltage, and \( I \) is the current.
04

Substitute Values

Substitute the given values into the formula. We have \( V = 1.41 \, \mathrm{V} \), \( E = 1.59 \, \mathrm{V} \), and \( I = 2 \, \mathrm{A} \). Thus, \( 1.41 = 1.59 - 2r \).
05

Solve for Internal Resistance

Rearrange the equation to solve for \( r \):\[ 1.59 - 1.41 = 2r \]Simplify:\[ 0.18 = 2r \]Divide by 2:\[ r = \frac{0.18}{2} = 0.09 \, \Omega \]
06

Verify the Solution

Plugging \( r = 0.09 \, \Omega \) back into the formula, we should match the original given voltages: \( 1.41 = 1.59 - 2 imes 0.09 \). The equation holds true, confirming our solution.

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in physics and electrical engineering. It describes the relationship between voltage, current, and resistance in an electrical circuit. The law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. It can be expressed with the formula:
  • \( V = I \times R \)
where:
  • \( V \) is the voltage (in volts)
  • \( I \) is the current (in amperes)
  • \( R \) is the resistance (in ohms)
Ohm's Law is crucial for understanding how electrical circuits behave. It helps us calculate unknown values when we know two of the three variables. In the context of internal resistance in cells, it assists in determining how much of the voltage is lost inside the cell due to its inherent resistance properties.

Understanding how Ohm's Law works is key to solving problems involving internal resistance. This allows us to focus on the voltage drop across the internal resistance when a known current flows through the cell.
open-circuit voltage
The open-circuit voltage of a cell is the voltage measured across the terminals when no current is flowing. It represents the maximum voltage available from a source when it is not connected to any load. In other words, it is the potential difference between the terminals when the circuit is open鈥攁 scenario with no external resistance applied.

The open-circuit voltage is crucial for determining the optimal performance of a battery. It reflects the energy a battery can supply, unaffected by any internal or external resistance losses. Usually, it is higher than the terminal voltage because it does not factor in any voltage drop due to the battery's internal resistance.

When analyzing circuits with batteries, knowing the open-circuit voltage helps in calculating the expected performance of the battery under load. It's important to remember that the open-circuit voltage alone isn't sufficient to determine the battery's effectiveness under load conditions; we also need to consider other factors, such as internal resistance, which affects the voltage when the battery supplies current.
terminal voltage
Terminal voltage is the actual voltage output of a power source when it is connected to a load and supplying current. It is the voltage across the terminals of a device like a battery when the circuit is closed, meaning there is current flow.

Sonied by the internal resistance of the power source:
  • The terminal voltage is less than the open-circuit voltage.
  • The difference is due to the voltage drop across the internal resistance.
In the given exercise, the terminal voltage is 1.41 V, which is lower than the open-circuit voltage of 1.59 V. This is because the internal resistance within the cell consumes part of the voltage as current flows through the cell.

Understanding terminal voltage is critical when assessing the actual performance and efficiency of an electrical power source. It illustrates that internal resistance decreases the effective voltage output available for work outside the battery, and calculations must account for such factors to predict performance accurately in a real-world scenario.

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Most popular questions from this chapter

In the Bohr model, the electron of a hydrogen atom moves in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\) with a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Determine its frequency \(f\) and the current \(I\) in the orbit. $$ f=\frac{v}{2 \pi r}=\frac{2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}=6.6 \times 10^{15} \mathrm{rev} / \mathrm{s} $$ Each time the electron goes around the orbit, it carries a charge \(e\) around the loop. The charge passing a point on the loop each second is $$ I=e f=\left(1.6 \times 10^{-19} \mathrm{C}\right)\left(6.6 \times 10^{15} \mathrm{~s}^{-1}\right)=1.06 \times 10^{-3} \mathrm{~A}=1.1 \mathrm{~mA} $$

A steady current of \(0.50 \mathrm{~A}\) flows through a wire. How much charge passes through the wire in one minute? Because \(I=q / t\), it follows that \(q=I t=(0.50 \mathrm{~A})(60 \mathrm{~s})=30 \mathrm{C} .(\) Recall that \(1 \mathrm{~A}=1 \mathrm{C} / \mathrm{s}\).)

An electron gun in a TV set shoots out a beam of electrons. The beam current is \(1.0 \times 10^{-5} \mathrm{~A}\). How many electrons strike the TV screen each second? How much charge strikes the screen in a minute?

(This problem introduces a unit sometimes used in the United States.) Number 24 copper wire has diameter \(0.0201\) in. Compute (a) the cross-sectional area of the wire in circular mils and (b) the resistance of \(100 \mathrm{ft}\) of the wire. The resistivity of copper is \(10.4 \Omega\).circular mils \(/ \mathrm{ft}\). The area of a circle in circular mils is defined as the square of the diameter of the circle expressed in mils, where \(1 \mathrm{mil}=0.001 \mathrm{in}\). (a) Area in circular mils \(=(20.1 \mathrm{mil})^{2}=404\) circular mils (b) \(\quad R=\rho \frac{L}{A}=\frac{(10.4 \Omega \cdot \text { circular } \mathrm{mil} / \mathrm{ft}) 100 \mathrm{ft}}{404 \text { circular mils }}=2.57 \Omega\)

A wire that has a resistance of \(5.0 \Omega\) is passed through an extruder so as to make it into a new wire three times as long as the original. What is the new resistance? Use \(R=\rho L / A\) to find the resistance of the new wire. To find \(\rho\), use the original data for the wire. Let \(L_{0}\) and \(A_{0}\) be the initial length and cross-sectional area, respectively. Then $$ 5.0 \Omega=\rho L_{0} / A_{0} \quad \text { or } \quad \rho=\left(A_{0} / L_{0}\right)(5.0 \Omega) $$ We were told that \(L=3 L_{0}\). To find \(A\) in terms of \(A_{0}\), note that the volume of the wire cannot change. Hence, $$ V_{0}=L_{0} A_{0} \quad \text { and } \quad V_{0}=L A $$ from which $$ L A=L_{0} A_{0} \quad \text { or } \quad A=\left(\frac{L_{0}}{L_{0}}\right)\left(A_{0}\right)=\frac{A_{0}}{3} $$ Therefore, \(\quad R=\frac{\rho L}{A}=\frac{\left(A_{0} / L_{0}\right)(5.0 \Omega)\left(3 L_{0}\right)}{A_{0} / 3}=9(5.0 \Omega)=45 \Omega\)
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