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Chapter 25: Problem 49

A \(25.0-\Omega\) bulb is connected across the terminals of a \(12.0-\mathrm{V}\) battery having 3.50\(\Omega\) of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

Short Answer

Expert verified
About 12.2% of the battery's power is lost in the internal resistance.

Step by step solution

01

Find the Total Resistance

To begin solving the problem, we need to find the total resistance in the circuit. This is the sum of the bulb's resistance and the battery's internal resistance, given as \( R_{total} = 25.0\, \Omega + 3.50\, \Omega = 28.5\, \Omega \).
02

Calculate the Total Current

Using Ohm's Law, calculate the current flowing through the circuit. Ohm's Law states \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. Rearranging for current, we have \( I = \frac{V}{R} \). Therefore, \( I = \frac{12.0\, V}{28.5\, \Omega} \approx 0.421\, A \).
03

Determine Power Dissipated in Internal Resistance

To find the power dissipated by the internal resistance, use the formula \( P = I^2R \). Substituting the values, we get \( P_{int} = (0.421\, A)^2 \times 3.5\, \Omega \approx 0.618\, W \).
04

Calculate the Total Power Outputted by the Battery

The total power supplied by the battery is given by \( P_{total} = VI \). So, \( P_{total} = 12.0\, V \times 0.421\, A \approx 5.052\, W \).
05

Find the Percentage of Power Lost to Internal Resistance

To find the percentage of power dissipated by the internal resistance, use the formula \( \frac{P_{int}}{P_{total}} \times 100\% \). Substituting the given power values, we find \( \frac{0.618\, W}{5.052\, W} \times 100\% \approx 12.2\% \).

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Key Concepts

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

Circuit Resistance
Circuit resistance is a fundamental concept in electrical circuits. It refers to the total opposition that a circuit presents to the flow of electric current. This resistance is expressed in Ohms (\( \Omega \)). In our example, the circuit resistance includes both the resistance of the bulb and the battery's internal resistance.

To calculate the total resistance in a circuit, simply add up all the individual resistances that current will pass through. In the given exercise, the bulb's resistance is \(25.0\, \Omega\) and the battery's internal resistance is \(3.50\, \Omega\). When combined, the circuit's total resistance is \(28.5\, \Omega\).

Knowing the total resistance is crucial because it affects how much current flows through the circuit. According to Ohm's Law, the current (\(I\)) is equal to the total voltage (\(V\)) divided by the total resistance (\(R\)):
  • \(I = \frac{V}{R}\)
Power Dissipation
Power dissipation refers to the loss of power (or energy) in a circuit, which is not utilized for any productive work. This lost power converts to heat due to the resistance in the wires or components of the circuit.

Power in a resistor can be calculated using the formula \(P = I^2R\), where \(P\) is the power in watts, \(I\) is the current in amperes, and \(R\) is the resistance in ohms. This tells us how much electrical power is turned into heat by the resistor. For our problem, the internal resistance dissipates part of the battery power as heat.

In general, the total power supplied by a battery in a circuit can be found using the equation:
  • \(P_{total} = VI\), where \(V\) is the voltage and \(I\) is the current.
The problem calculates that, out of the total power supplied by the battery, some percentage is lost to power dissipation within the battery's internal resistance. Calculating this helps in understanding energy efficiency in circuits.
Internal Resistance
Internal resistance is an inherent characteristic of all batteries and power sources. It represents the resistance to the flow of current within the battery itself, often caused by ions moving through an electrolyte solution inside the battery.

This internal resistance reduces the voltage available to the external circuit, as some energy is consumed overcoming this resistance. Thus, it results in power loss, termed as 'internal power dissipation'. In the exercise provided, the internal resistance of the battery is given as \(3.50\, \Omega\).

The power dissipated internally is an important factor because it determines how efficiently a battery uses its energy to power external components, such as the bulb in our example. If a large portion of power is lost internally, less power remains available to perform the function needed in a circuit.

Understanding internal resistance helps in designing better power systems, allowing more effective transmission and usage of electrical energy in appliances, thus enhancing overall system efficiency.

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Most popular questions from this chapter

A \(1.50-\mathrm{m}\) cylinder of radius 1.10 \(\mathrm{cm}\) is made of a complicated mixture of materials. Its resistivity depends on the distance \(x\) from the left end and obeys the formula \(\rho(x)=\) \(a+b x^{2},\) where \(a\) and \(b\) are constants. At the left end, the resistivity is \(2.25 \times 10^{-8} \Omega \cdot \mathrm{m},\) while at the right end it is \(8.50 \times\) \(10^{-8} \Omega \cdot \mathrm{m}\) . (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries a \(1.75-\) A current? (c) If we cut the rod into two 75.0 -cm halves, what is the resistance of each half?

A tightly coiled spring having 75 coils, each 3.50 \(\mathrm{cm}\) in diameter, is made of insulated metal wire 3.25 \(\mathrm{mm}\) in diameter. An ohmmeter connected across its opposite ends reads 1.74\(\Omega\) ? What is the resistivity of the metal?

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length \(L\) and cross- sectional area \(A\) lies along the \(x\) -axis between \(x=0\) and \(x=L .\) The material obeys Ohm's law, and its resistivity varies along the rod according to \(\rho(x)=\) \(\rho_{0} \exp (-x / L) .\) The end of the rod at \(x=0\) is at a potential \(V_{0}\) greater than the end at \(x=L .\) (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude \(E(x)\) in the rod as a function of \(x .\) (c) Find the electric potential \(V(x)\) in the rod as a function of \(x .\) (d) Graph the functions \(\rho(x), E(x),\) and \(V(x)\) for values of \(x\) between \(x=0\) and \(x=L\) .

A \(12.0-\mathrm{V}\) battery has an internal resistance of 0.24 \(\mathrm{s}\) and a capacity of 50.0 \(\mathrm{A} \cdot \mathrm{h}\) (see Exercise 25.47\() .\) The battery is charged by passing a 10 -A current through it for 5.0 \(\mathrm{h}\) . (a) What is the terminal voltage during charging? (b) What total electricalenergy is supplied to the battery during charging? (c) What electrical energy is dissipated in the internal resistance during charging? (d) The battery is now completely discharged through a resistor, again with a constant current of 10 \(\mathrm{A}\) . What is the external circuit resistance? (e) What total electrical energy is supplied to the external resistor? (f) What total electrical energy is dissipated in the internal resistance? (g) Why are the answers to parts (b) and (e) not the same?

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads 12.6 \(\mathrm{V} .\) You cut off a \(20.0-\mathrm{m}\) length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A. You then cut off a \(40.0-\mathrm{m}\) length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.20 A. Even though the equipment you have available to you is limited, your boss assures you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?
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