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Chapter 19: Problem 26

A 6.00 \(\mathrm{V}\) lantern battery is connected to a 10.5\(\Omega\) lightbulb, and the resulting current in the circuit is 0.350 A. What is the internal resistance of the battery?

Short Answer

Expert verified
The internal resistance of the battery is approximately 6.643 惟.

Step by step solution

01

Understand Ohm's Law

Ohm's Law is essential for analyzing circuits, and it states that the voltage across a resistor is the product of the current flowing through it and its resistance: \( V = I \times R \).
02

Identify Known Values

We are given that the voltage of the battery \( V = 6.00 \, \mathrm{V} \), the resistance of the lightbulb \( R = 10.5 \, \Omega \), and the current \( I = 0.350 \, \mathrm{A} \).
03

Calculate Total Voltage across the Lightbulb

Using Ohm's Law, calculate the voltage across the lightbulb. This is \( V = I \times R = 0.350 \, \mathrm{A} \times 10.5 \, \Omega = 3.675 \, \mathrm{V} \).
04

Calculate Lost Voltage in the Battery

The lost voltage, or voltage drop due to the internal resistance of the battery, is the difference between the total battery voltage and the voltage across the lightbulb: \( 6.00 \, \mathrm{V} - 3.675 \, \mathrm{V} = 2.325 \, \mathrm{V} \).
05

Determine the Internal Resistance

Using Ohm's Law again, the internal resistance \( r \) is calculated by \( r = \frac{\text{Lost Voltage}}{I} = \frac{2.325 \, \mathrm{V}}{0.350 \, \mathrm{A}} = 6.643 \, \Omega \).

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Key Concepts

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

Circuit Analysis
Circuit analysis is a fundamental skill in understanding how electric circuits work. Essentially, this process involves determining how different components like resistors, batteries, and lightbulbs interact within a circuit to influence the overall behavior.
A basic principle we rely on for circuit analysis is **Ohm's Law**, which can be stated as:
  • Formula: \( V = I \times R \)
  • Where, \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
This law helps us find unknown values by rearranging to solve for the desired variable.
It also helps us understand how voltage is distributed in a circuit, allowing us to calculate factors like the provided voltage from a battery compared to what is used by a resistive component like a lightbulb. Recognizing these relationships is crucial for accurately analyzing any circuit.
Internal Resistance
The concept of internal resistance is pivotal in determining how effective a battery is within a circuit. Internal resistance refers to the resistance within the battery itself, which can cause energy loss as heat and reduce the effective voltage delivered to an external load.
In analyzing circuits, understanding internal resistance allows us to determine:
  • How much voltage is dropped simply within the battery.
  • How this affects the overall current supply to various elements within the circuit.
For example, using Ohm's Law, we can calculate the internal resistance \( r \) using the formula:\[ r = \frac{\text{Lost Voltage}}{I} \]This is derived from knowing both the voltage drop due to internal resistance and the current flowing from the battery, which helps us find how much resistance is affecting the circuit internally. Keeping track of and minimizing internal resistance can improve battery performance.
Voltage Drop
The voltage drop is a concept that refers to the reduction in voltage as electric current flows through a component in a circuit. It is an important factor to consider because it indicates how much energy is used by a specific component, and thus how much is left for other parts of the circuit.
There are two major kinds of voltage drops in any basic circuit:
  • Voltage drop across external components: This occurs through elements like resistors or bulbs, calculated using Ohm's Law \( V = I \times R \).
  • Voltage drop due to internal resistance: This occurs within the battery or power source, reflecting the energy dissipated inside the battery itself.
Understanding these drops is key to determining how much power is effectively being supplied to a circuit relative to what's being generated by the power source. By accounting for voltage drops, we ensure that circuits are efficiently designed and that components receive the power they need to function properly.

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Most popular questions from this chapter

A \(540-\mathrm{W}\) electric heater is designed to operate from 120 \(\mathrm{V}\) lines. (a) What is its resistance? (b) What current does it draw? (c) If the line voltage drops to \(110 \mathrm{V},\) what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

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?

Three resistors having resistances of \(1.60 \Omega, 2.40 \Omega,\) and \(4.80 \Omega,\) respectively, are connected in parallel to a 28.0 \(\mathrm{V}\) bat- tery that has negligible internal resistance. Find (a) the equivalent resistance of the combination, (b) the current in each resistor, (c) the total current through the battery, (d) the voltage across each resistor, and (e) the power dissipated in each resistor. (f) Which resistor dissipates the most power, the one with the greatest resistance or the one with the least resistance? Explain why this should be.

Electric eels. Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500 \(\mathrm{V}\) and produce currents of 80 \(\mathrm{mA}\) (or even larger). A typical pulse lasts for 10 \(\mathrm{ms}\) . What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

If you triple the length of a cable and at the same time double its diameter, what will be its resistance if its original resistance was \(R\) ?
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