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A resistor has a resistance \(R\), and a battery has an internal resistance \(r\). When the resistor is connected across the battery, ten percent less power is dissipated in \(R\) than would be dissipated if the battery had no internal resistance. Find the ratio \(r / R\).

Short Answer

Expert verified
The ratio \( r/R \) is \( \frac{1 - \sqrt{0.9}}{\sqrt{0.9}} \).

Step by step solution

01

Determine Power with No Internal Resistance

When there is no internal resistance, the voltage across the resistor is equal to the battery's electromotive force (emf), let's denote it as \( V \). The power dissipated in the resistor \( R \) can be calculated by the formula \( P_0 = \frac{V^2}{R} \).
02

Determine Power with Internal Resistance

With internal resistance \( r \), the effective voltage across the resistor is reduced. The total resistance in the circuit is \( R + r \). The voltage across \( R \) is \( V' = \frac{V imes R}{R + r} \). The power dissipated in \( R \) is \( P = \frac{(V')^2}{R} = \frac{V^2 imes R}{(R + r)^2} \).
03

Set Up the Equation for Ten Percent Less Power

We are given that \( P \) is ten percent less than \( P_0 \). This gives us the equation: \( \frac{V^2 imes R}{(R + r)^2} = 0.9 \times \frac{V^2}{R} \).
04

Simplify the Equation

Cancel out \( V^2 \) from both sides and rearrange to get: \( \frac{R^2}{(R + r)^2} = 0.9 \).
05

Solve for the Ratio r/R

Take the square root of both sides to obtain \( \frac{R}{R + r} = \sqrt{0.9} \). Solve for \( r \) in terms of \( R \): \( r = R \left( \frac{1}{\sqrt{0.9}} - 1 \right) \). Simplify this to find \( \frac{r}{R} = \frac{1 - \sqrt{0.9}}{\sqrt{0.9}} \).

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Key Concepts

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

Internal Resistance
Internal resistance is a key factor in understanding the behavior of real-life batteries. It refers to the resistance within the battery itself, which can impede the flow of current. This internal resistance causes the terminal voltage of the battery to be lower than its electromotive force (emf) when current flows through the circuit.
When you connect a resistor to a battery, two types of resistances are at play:
  • The resistance of the resistor itself, denoted as \( R \).
  • The internal resistance of the battery, denoted as \( r \).
The presence of internal resistance leads to a decrease in the voltage available across the external resistor, reducing the power it can dissipate. Understanding this concept is essential for accurate calculations in circuits involving power sources.
Electromotive Force (emf)
Electromotive force (emf) is a fundamental concept in electrical circuits. Contrary to what its name suggests, it is not actually a "force". Instead, it is the voltage generated by a battery or other power source when no current is flowing. Think of it as the maximum potential difference the source can provide.
The emf is often denoted by \( V \), representing the ideal voltage of the power source without any load. However, once the current starts to flow and the internal resistance comes into play, the effective voltage (or terminal voltage) seen by the circuit is reduced. This reduction is crucial to consider when analyzing circuits as it affects the performance and efficiency of the power source, especially in practical applications.
Circuit Voltage
Circuit voltage, particularly in the context of this problem, refers to the actual voltage across the external resistor when connected to a battery with both resistance and internal resistance. It's not the same as the emf. Instead, it's the voltage after accounting for the voltage drop due to the internal resistance.
The voltage drop across internal resistance can be calculated using Ohm's Law. In our problem, this drop reduces the voltage from its emf value to a lower effective voltage \( V' \) which is given by:
  • \( V' = \frac{V \times R}{R + r} \),
where \( V \) is the emf, \( R \) is the resistor's resistance, and \( r \) is the internal resistance. This effective voltage is crucial to determining the actual power being dissipated in the real-world scenario.
Resistance Ratio
The resistance ratio \( \frac{r}{R} \) is an expression that compares the internal resistance of the battery to the resistance of the external resistor. This ratio is significant as it directly influences the circuit's efficiency and performance.
In the provided exercise, we determined this ratio using the fact that the power dissipated across the resistor with internal resistance is ten percent less than if there were no internal resistance at all. By setting up and solving the equation for the power difference, we deduced that the ratio can be expressed as:
  • \( \frac{r}{R} = \frac{1 - \sqrt{0.9}}{\sqrt{0.9}} \)
This expression illustrates the impact of internal resistance on the circuit's overall power dissipation. Understanding and calculating resistance ratios are essential for designing efficient electrical systems where power loss needs to be minimized.

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Most popular questions from this chapter

The total current delivered to a number of devices connected in parallel is the sum of the individual currents in each device. Circuit breakers are resettable automatic switches that protect against a dangerously large total current by "opening" to stop the current at a specified safe value. A 1650 -W toaster, a 1090 -W iron, and a 1250 -W microwave oven are turned on in a kitchen. As the drawing shows, they are all connected through a 20 -A circuit breaker to an ac voltage of \(120 \mathrm{~V}\). (a) Find the equivalent resistance of the three devices. (b) Obtain the total current delivered by the source and determine whether the breaker will "open" to prevent an accident.

A resistor (resistance \(=R\) ) is connected first in parallel and then in series with a \(2.00-\Omega\) resistor. A battery delivers five times as much current to the parallel combination as it does to the series combination. Determine the two possible values for \(R\).

An electric furnace runs nine hours a day to heat a house during January ( 31 days). The heating element has a resistance of \(5.3 \Omega\) and carries a current of \(25 \mathrm{~A}\). The cost of electricity is \(\$ 0.12 / \mathrm{kWh}\). Find the cost of running the furnace for the month of January.

An electronic flash attachment for a camera produces a flash by using the energy stored in a \(150-\mu \mathrm{F}\) capacitor. Between flashes, the capacitor recharges through a resistor whose resistance is chosen so the capacitor recharges with a time constant of \(3.0 \mathrm{~s}\). Determine the value of the resistance.

The recovery time of a hot water heater is the time required to heat all the water in the unit to the desired temperature. Suppose that a 52-gal \(\left(1.00 \mathrm{gal}=3.79 \times 10^{-3} \mathrm{~m}^{3}\right)\) unit starts with cold water at \(11^{\circ} \mathrm{C}\) and delivers hot water at \(53^{\circ} \mathrm{C}\). The unit is electric and utilizes a resistance heater \((120 \mathrm{~V} \mathrm{ac}, 3.0 \Omega)\) to heat the water. Assuming that no heat is lost to the environment, determine the recovery time (in hours) of the unit.

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