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Chapter 26: Problem 32

The internal resistance of a \(6.4-\mathrm{V}\) storage battery is \(4.8 \mathrm{~m} \Omega\). What is the theoretical maximum current on short circuit? (In practice the leads and connections have some resistance, and this theoretical value would not be attained.)

Short Answer

Expert verified
The theoretical maximum current on short circuit is approximately 1333.33 A.

Step by step solution

01

Understand the Problem

We must find the theoretical maximum current that can flow when the voltage source (battery) is short-circuited. The internal resistance of the battery is given, along with its voltage rating.
02

Apply Ohm's Law

Ohm鈥檚 Law is represented as \( I = \frac{V}{R} \). For a short circuit, the load resistance is zero, so the only resistance is the battery's internal resistance \( R = 4.8 \, \text{m}\Omega \).
03

Calculate the Theoretical Maximum Current

Use \( I = \frac{V}{R} \) with \( V = 6.4 \, \text{V} \) and \( R = 4.8 \, \text{m}\Omega \). Thus, \( I = \frac{6.4}{4.8 \times 10^{-3}} = \frac{6.4}{0.0048} \approx 1333.33 \, \text{A} \).

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Key Concepts

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

Internal Resistance
Internal resistance refers to the opposition within a battery or other power source to the flow of electric current. It arises due to the materials and construction of the battery itself. This resistance causes some of the energy produced by the battery to be lost as heat.
  • It is typically measured in ohms (惟).
  • In our example, the battery's internal resistance is given as 4.8 milliohms (m惟).
This concept is important because it affects how much current a battery can deliver. The higher the internal resistance, the less efficient the battery is in supplying current. Especially in high-demand situations, such as a short circuit, internal resistance limits the current, thus ensuring the battery doesn't deliver infinite current.
Maximum Current
The maximum current a battery can supply under specific conditions is determined by both its voltage and its internal resistance. According to Ohm's Law:\[ I = \frac{V}{R} \]where:
  • \( I \) is the current (in amperes),
  • \( V \) is the voltage (in volts),
  • \( R \) is the resistance (in ohms).
In the case of the exercise, the battery voltage is 6.4 volts, and the internal resistance is 4.8 m惟. This gives us:\[ I = \frac{6.4}{4.8 \times 10^{-3}} \]Calculating this gives a theoretical value of approximately 1333.33 amperes for the maximum current. This value represents the largest amount of current that can flow through the battery given its internal characteristics, without considering any other external resistance.
Short Circuit
A short circuit occurs when a path is created that allows a large current to flow with very little or no resistance. In simple terms, it's like creating a shortcut for electricity. This is dangerous because it can lead to:
  • Overheating due to excessive current,
  • Damage to electrical components,
  • Potential fires or explosions.
In the theoretical scenario given, when a battery is short-circuited, the internal resistance is the only resistance. Because no other load is present, the current is only limited by this internal resistance as calculated before. This helps us understand why internal resistance plays such a crucial role in determining the safety and efficiency of electrical systems.

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Most popular questions from this chapter

A coil of wire has a resistance of \(25.00 \Omega\) at \(20^{\circ} \mathrm{C}\) and a resistance of \(25.17 \Omega\) at \(35^{\circ} \mathrm{C}\). What is its temperature coefficient of resistance?

Number 10 wire has a diameter of \(2.59 \mathrm{~mm}\). How many meters of number 10 aluminum wire are needed to give a resistance of \(1.0 \Omega\) ? \(\rho\) for aluminum is \(2.8 \times 10^{-8} \Omega \cdot \mathrm{m}\). From \(R=\rho L / A\), $$ L=\frac{R A}{\rho}=\frac{(1.0 \Omega)(\pi)\left(2.59 \times 10^{-3} \mathrm{~m}\right)^{2} / 4}{2.8 \times 10^{-8} \Omega \cdot \mathrm{m}}=0.19 \mathrm{~km} $$

(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) \(\quad R=\rho \frac{L}{A}=\frac{(10.4 \Omega \cdot \text { circular } \mathrm{mil} / \mathrm{ft}) 100 \mathrm{ft}}{404 \text { circular mils }}=2.57 \Omega\)

How many electrons per second pass through a section of wire carrying a current of \(0.70 \mathrm{~A}\) ?

A metal rod is \(2 \mathrm{~m}\) long and \(8 \mathrm{~mm}\) in diameter. Compute its resistance if the resistivity of the metal is \(1.76 \times 10^{-8} \Omega \cdot \mathrm{m}\). $$ R=\rho \frac{L}{A}=\left(1.76 \times 10^{-8} \Omega \cdot \mathrm{m}\right) \frac{2 \mathrm{~m}}{\pi\left(4 \times 10^{-3} \mathrm{~m}\right)^{2}}=7 \times 10^{-4} \Omega $$
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