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Chapter 25: Problem 46

A typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.

Short Answer

Expert verified
The solutions are: (a) The power delivered to the bulb is 0.529 W. (b) The total energy delivered to the bulb is 9522 J. (c) The combined internal resistance of the batteries when the power decreases to half of its initial value is 10 Ohm.

Step by step solution

01

Calculate the power delivered to the bulb (a)

The total emf from the batteries is 2*1.5V = 3.0V, since they are arranged in series. The resistance for the bulb is given as 17 Ohm. By Ohm's law, we find the current in the circuit: I = V/R = 3.0V / 17 Ohm = 0.176 A. To find the power delivered to the bulb, we use the formula P = IV = 0.176 A * 3.0V = 0.529 W.
02

Calculate the total energy delivered to the bulb (b)

To find the total energy delivered to the bulb, we use the equation E = Pt. This gives us E = 0.529 W * 5.0 h * 3600 s/h = 9522 J.
03

Find the combined internal resistance of both batteries (c)

When the power delivered to the bulb is halved, the power is 0.529W / 2 = 0.2645W. We can use P=I^2R to find the new current I which is sqrt(P/R)=sqrt(0.2645W/17 Ohm) = 0.111A. Then by using Ohm's law (V=IR) we can find the total resistance in the circuit Rt which is V/I = 3.0V/0.111A = 27 Ohm. The combined internal resistance of the batteries is therefore Rt - Rbulb = 27 Ohm - 17 Ohm = 10 Ohm.

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Key Concepts

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

Electric Circuits
Electric circuits are pathways that allow electricity to flow, enabling the operation of many electronic devices and systems. These circuits typically include components such as resistors, capacitors, and power sources like batteries. In the given exercise, we have a simple circuit composed of batteries connected in series to a bulb.
  • Components in series share the same current but the voltage across each component is different.
  • In this setup, the circuit is completed, enabling current to flow from the battery through the bulb and back to the battery.
Understanding electric circuits is crucial because they are the backbone of most electrical and electronic devices. Recognizing how components interact within a circuit helps in diagnosing issues and optimizing performance.
Ohm's Law
Ohm's Law is a fundamental principle in physics, describing the relationship between voltage, current, and resistance in an electric circuit. It is expressed by the equation: \( V = IR \), where:
  • \( V \) is the voltage across the component (in volts).
  • \( I \) is the current flowing through the component (in amperes).
  • \( R \) is the resistance of the component (in ohms).
In our flashlight problem, Ohm's Law is applied to find the current flowing through the bulb. With a total voltage (from the series batteries) equal to 3.0V and a bulb resistance of 17\( \Omega \), the law enables us to calculate the current:\( I = \frac{V}{R} = \frac{3.0}{17} \approx 0.176 \) A.Ohm鈥檚 Law is essential for designing and understanding circuits, allowing us to predict how changes in voltage or resistance will affect the current.
Power Calculation
Power in an electric circuit refers to the rate at which energy is used or transferred by a component. In circuits, power can be calculated using the formula:\( P = IV \)Where:
  • \( P \) is the power (in watts).
  • \( I \) is the current (in amperes).
  • \( V \) is the voltage (in volts).
For the flashlight's bulb, the power can be calculated as:\( P = 0.176 \text{ A} \times 3.0 \text{ V} = 0.529 \text{ W} \).Power calculations are essential for ensuring that components like bulbs function properly without exceeding their power ratings, which could cause damage or inefficient operation.
Internal Resistance
Internal resistance is a characteristic of all power sources, such as batteries, that impedes the flow of electric current. The resistance comes from the material within the battery and affects how much voltage is available for external circuits like a bulb in the flashlight.
  • High internal resistance can lower the efficiency of power delivery to a circuit.
  • Over time, as batteries discharge, their internal resistance tends to increase.
In the flashlight example, as the batteries age or their charge depletes, internal resistance increases. This can be calculated by analyzing how power output diminishes; in this exercise, the internal resistance needed to reduce the bulb's power by half was found to be 10\( \Omega \). Understanding internal resistance is crucial for designing circuits with efficient energy delivery and longevity.
Energy Delivery
The concept of energy delivery in circuits refers to the amount of energy transferred from a power source to a load over time. The energy delivered can be determined using the formula:\( E = Pt \)Where:
  • \( E \) is the energy (in joules).
  • \( P \) is the power (in watts).
  • \( t \) is the time (in seconds).
In this exercise, we calculated energy delivered to the flashlight's bulb over a period of 5 hours. Using the previous power calculation of 0.529 W and multiplying it by the time in seconds, we find:\( E = 0.529 \text{ W} \times 5.0 \times 3600 \) s = 9522 J.Understanding energy delivery helps in planning for battery usage, ensuring that devices operate for the desired duration, and managing energy consumption efficiently.

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Most popular questions from this chapter

A battery has emf \(30.0 \mathrm{~V}\) and internal resistance \(r .\) A \(9.00 \Omega\) resistor is connected to the terminals of the battery, and the voltage drop across the resistor is \(27.0 \mathrm{~V}\). What is the internal resistance of the battery?

Two cylindrical cans with insulating sides and conducting end caps are filled with water, attached to the circuitry shown in Fig. \(\mathbf{P} 25.69,\) and used to determine salinity levels. The cans are identical, with radius \(r=5.00 \mathrm{~cm}\) and length \(L=3.00 \mathrm{~cm} .\) The battery supplies a potential of \(10.0 \mathrm{~V},\) has a negligible internal resistance, and is connected in series with a resistor \(R=15.0 \Omega .\) The left cylinder is filled with pure distilled water, which has infinite resistivity. The right cylinder is filled with a saltwater solution. It is known that the resistivity of the saltwater solution is determined by the relationship \(\rho=\left(s_{0} / s\right) \Omega \cdot \mathrm{m},\) where \(s\) is the salinity in parts per thousand \((\mathrm{ppt})\) and \(s_{0}=6.30\) ppt. (a) The ammeter registers a current of \(484 \mathrm{~mA}\). What is the salinity of the saltwater solution? (b) The left cylinder acts as a capacitor. Use Eq. (24.19) for its capacitance. How much charge is present on its upper plate? Note that pure water has a dielectric constant of \(80.4 .\) (c) At what rate is energy dissipated by the saltwater? (d) For what salinity level would the \(15.0 \Omega\) resistor dissipate half the power supplied by the battery?

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 assure 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?

Lightning Strikes. During lightning strikes from a cloud to the ground, currents as high as 25,000 A can occur and last for about \(40 \mu\) s. How much charge is transferred from the cloud to the earth during such a strike?

A material with resistivity \(\rho\) is formed into a cylinder of length \(L\) and outer radius \(r_{\text {outer }}\). A cylindrical core with radius \(r_{\text {inner }}\) is removed from the axis of this cylinder and filled with a conducting material, which is attached to a wire. The outer surface of the cylinder is coated with a conducting material and attached to another wire. (a) If the second wire has potential \(V\) greater than the first wire, in what direction does the local electric field point inside of the cylinder? (b) The magnitude of this electric field is \(c / r,\) where \(c\) is a constant and \(r\) is the distance from the axis of the cylinder. Use the relationship \(V=\int \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{l}}\) to determine the constant \(c .(\mathrm{c})\) What is the resistance of this device? (d) A \(1.00-\mathrm{cm}\) -long hollow cylindrical resistor has an inner radius of \(1.50 \mathrm{~mm}\) and an outer radius of \(3.00 \mathrm{~mm} .\) The material is a blend of powdered carbon and ceramic whose resistivity \(\rho\) may be altered by changing the amount of carbon. If this device should have a resistance of \(6.80 \mathrm{k} \Omega,\) what value of \(\rho\) should be selected?
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