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Chapter 25: Problem 52

An idealized voltmeter is connected across the terminals of a \(15.0-\mathrm{V}\) battery, and a \(75.0-\Omega\) appliance is also connected across its terminals. If the voltmeter reads \(11.3 \mathrm{V} :\) (a) how much power is being dissipated by the appliance, and \((b)\) what is the internal resistance of the battery?

Short Answer

Expert verified
(a) Power dissipated: 1.70 W; (b) Internal resistance: 24.54 惟.

Step by step solution

01

Understanding the Scenario

We have a battery with terminals providing 15.0 V when disconnected but only 11.3 V when connected to a voltmeter and a 75.0-鈩� appliance. The voltage drop indicates the presence of internal resistance in the battery.
02

Calculate Current Through the Appliance

Using Ohm's Law, \( V = IR \), where \( V = 11.3 \text{ V} \) and \( R = 75.0 \Omega \), we find the current, \( I \). Thus, \( I = \frac{11.3}{75.0} \approx 0.1507 \text{ A} \).
03

Calculate Power Dissipated by the Appliance

Power, \( P \), dissipated by the appliance is calculated using the formula \( P = I^2 R \). Substituting the known values, \( P = (0.1507)^2 \times 75.0 \approx 1.70 \text{ W} \).
04

Calculate Internal Resistance of the Battery

The internal resistance \( r \) can be found using the formula for the total voltage, \( V = \varepsilon - Ir \), where \( \varepsilon = 15.0 \text{ V} \) and \( V = 11.3 \text{ V} \). Substituting in, we have \( 15.0 - 11.3 = 0.1507 \times r \). Solving for \( r \), we have \( 3.7 = 0.1507 \times r \), so \( r \approx 24.54 \Omega \).

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Key Concepts

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

Power Dissipation in Electrical Circuits
Power dissipation in electrical circuits refers to the process by which electrical energy is converted into other forms of energy, such as heat, more due to the resistance in the circuit. When a current flows through a resistor, the resistor tends to dissipate energy in the form of heat. This phenomenon can be quantitatively described using the power dissipation formula:
  • \( P = I^2 R \)
where \( P \) is the power dissipated in watts, \( I \) is the current in amperes, and \( R \) is the resistance in ohms.
In the given exercise, we calculated the power dissipated by the appliance when connected to the battery. With the current determined as approximately 0.1507 A and the resistance as 75.0 \( \Omega \), the power dissipation was found to be around 1.70 W. This means that the appliance converts 1.70 joules of electrical energy into heat every second.
This concept is vital when assessing the thermal performance and energy efficiency of electrical devices and helping to prevent overheating in circuits.
Understanding Internal Resistance of a Battery
Internal resistance is the resistance that exists within the battery itself, causing a decrease in the terminal voltage when a current flows through it. This internal factor is not ideal because it reduces the efficiency of the battery by dissipating some of the energy as heat rather than supplying it fully to the connected load.In the context of the original exercise, the internal resistance (\( r \)) of the battery led to a terminal voltage of 11.3 V, even though the battery's electromotive force (\( \varepsilon \)) is 15 V. We used the relationship between electromotive force and terminal voltage:
  • \( V = \varepsilon - Ir \)
to solve for the internal resistance. By replacing the known values and solving the equation, we found the internal resistance to be 24.54 \( \Omega \).
Understanding internal resistance is crucial when designing and selecting batteries for certain tasks. It helps in calculating the effective efficiency and performance of the battery under load.
Characteristics of an Ideal Voltmeter
An ideal voltmeter is a theoretical instrument used for measuring electrical potential difference between two points in a circuit without affecting the circuit conditions. The key feature of an ideal voltmeter is its infinite internal resistance. This is to ensure that it draws zero current from the circuit it is measuring. If a voltmeter drew current, it would alter the circuit parameters, leading to inaccurate measurements. However, in real scenarios, it is impossible to achieve true ideality. Still, designing voltmeters with very high resistance makes them very close to ideal.
In our exercise scenario, the idealized voltmeter measured a reduced terminal voltage of 11.3 V. It indicates that the effect of the circuit鈥檚 internal structure was observed without the measurement tool further influencing the results. Even if theoretically ideal in this exercise, real-life voltmeters endeavor to approximate high resistance to approach this ideal behavior.

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Most popular questions from this chapter

A \(12.6-\mathrm{V}\) car battery with negligible internal resistance is connected to a series combination of a \(3.2-\Omega\) resistor that obeys Ohm's law and a thermistor that does not obey Ohm's law but instead has a current- voltage relationship \(V=\alpha I+\beta I^{2},\) with \(\alpha=3.8 \Omega\) and \(\beta=1.3 \Omega / \mathrm{A}\) . What is the current through the \(3.2-\Omega\) resistor?

The following measurements of current and potential difference were made on a resistor constructed of Nichrome wire: $$ \begin{array}{l|llll}{\mathbf{I}(\mathbf{A})} & {0.50} & {1.00} & {2.00} & {4.00} \\ {V_{a b}(\mathbf{v})} & {1.94} & {3.88} & {7.76} & {15.52}\end{array} $$ (a) Graph \(V_{a b}\) as a function of \(I .\) (b) Does Nichrome obey Ohm's law? How can you tell? (c) What is the resistance of the resistor in ohms?

A cylindrical copper cable 1.50 \(\mathrm{km}\) long is connected across a 220.0 - V potential difference. (a) What should be its diameter so that it produces heat at a rate of 50.0 \(\mathrm{W}\) ? (b) What is the electric field inside the cable under these conditions?

A current of 3.6 A flows through an automobile headlight. How many coulombs of charge flow through the beadlight in 3.0 \(\mathrm{h} ?\)

The Tolman-Stewart experiment in 1916 demonstrated that the free charges in a metal have negative charge and provided a quantitative measurement of their charge-to-mass ratio, \(|q| / m\) . The experiment consisted of abruptly stopping a rapidly rotating spool of wire and measuring the potential difference that this produced between the ends of the wire. In a simplified model of this experment, consider a metal rod of length \(L\) that is given a uniform acceleration \(\overrightarrow{\boldsymbol{d}}\) to the right. Initially the free charges in the metal lag behind the rod's motion, thus setting up an electric field \(\overrightarrow{\boldsymbol{E}}\) in the rod. In the steady state this field exerts a force on the free charges that makes them accelerate along with the rod. (a) Apply \(\Sigma \vec{F}=m \vec{d}\) to the free charges to obtain an expression for \(|q| / m\) in terms of the magnitudes of the induced electric field \(\overrightarrow{\boldsymbol{k}}\) and the acceleration \(\overrightarrow{\boldsymbol{d}} .\) (b) If all the free charges in the metal rod have the same acceleration, the electric field \(\overrightarrow{\boldsymbol{E}}\) is the same at all points in the rod. Use this fact to rewrite the expression for \(|q| / m\) in terms of the potential \(V_{b c}\) between the ends of the rod (Fig. 25.44\()\) . (c) If the free charges have negative charge, which end of the rod, \(b\) or \(c,\) is at higher potential? (d) If the rod is 0.50 \(\mathrm{m}\) long and the free charges are electrons (charge \(q=-1.60 \times 10^{-19} \mathrm{C},\) mass \(9.11 \times 10^{-31} \mathrm{kg} ),\) what magnitude of acceleration is required to produce a potential difference of 1.0 \(\mathrm{mV}\) between the ends of the rod?(e) Discuss why the actual experiment used a rotating spool of thin wire rather than a moving bar as in our simplified analysis.
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