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Chapter 25: Problem 61

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V. (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

Short Answer

Expert verified
Internal resistance is 0.36 ohms, and emf is 8.94 volts.

Step by step solution

01

Understand the Problem

We have a battery with two scenarios of potential difference and current. We need to find the internal resistance and the electromotive force (emf) of the battery. Use the formula that links these concepts: \[ V = \mathcal{E} - Ir \] for Scenario 1, and \[ V = \mathcal{E} + Ir \] for Scenario 2.
02

Write Equations for Each Scenario

For Scenario 1, with potential difference 8.40 V and current 1.50 A:\[ 8.40 = \mathcal{E} - 1.50r \]For Scenario 2, with potential difference 10.20 V and current 3.50 A (reverse direction, meaning potential difference increases):\[ 10.20 = \mathcal{E} + 3.50r \]
03

Solve for Internal Resistance (r)

Subtract the first equation from the second to eliminate \(\mathcal{E}\):\[ (10.20 - 8.40) = (\mathcal{E} + 3.50r) - (\mathcal{E} - 1.50r) \]\[ 1.80 = 5r \]Solve for \( r \):\[ r = \frac{1.80}{5} = 0.36 \text{ ohms} \]
04

Solve for Emf (\(\mathcal{E}\))

Use the value of \( r = 0.36 \) ohms in one of the original equations to find \(\mathcal{E}\). We'll use the first equation:\[ 8.40 = \mathcal{E} - 1.50(0.36) \]\[ 8.40 = \mathcal{E} - 0.54 \]Solve for \(\mathcal{E}\):\[ \mathcal{E} = 8.40 + 0.54 = 8.94 \text{ volts} \]
05

Compile the Solution

The internal resistance of the battery is 0.36 ohms, and the emf of the battery is 8.94 volts.

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Key Concepts

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

Electromotive Force (emf)
Electromotive Force, or emf, is a crucial concept in understanding how batteries, or cells, function within an electrical circuit. It represents the energy provided by the battery per coulomb of charge. Unlike the potential difference, which can vary, the emf is considered a constant value representing the maximum potential that the source can provide.
In essence, emf is the ability of a battery or generator to push electrons around a circuit, akin to the force that makes a pump move water through pipes. This is why it's often labeled as a force鈥攅ven though it technically isn't one in the traditional physics sense.
  • Measured in volts (V), emf is denoted by the symbol \( \mathcal{E} \).
  • It's calculated using the equation: \( \mathcal{E} = V + Ir \), where \( V \) is the terminal voltage, \( I \) is the current, and \( r \) is the internal resistance.
  • In the context of the problem, the emf tells us how much voltage the battery can ideally provide without any losses.
  • Using our calculated internal resistance \( r = 0.36 \text{ ohms} \), the emf can be found by rearranging and solving the formula with given potential differences and currents.
Potential Difference
Potential difference, commonly known as voltage, is the amount of energy needed to move a charge from one point to another in a circuit. It's the key reason why electrons move through a wire, leading to the flow of electricity.
In practical terms, potential difference is what causes an electrical appliance to work when connected to a power source. It can be thought of as the pressure that pushes charges through a circuit, analogous to how water pressure pushes water through a pipe.
  • The potential difference is measured in volts (V).
  • It's often denoted by the symbol \( V \), but don't confuse it with emf, which is the maximum potential difference provided by a battery.
  • In our original problem, potential difference values change when the direction and magnitude of the current change, stating values of 8.40 V and 10.20 V.
  • These changes depend on how the internal resistance of the battery affects the flow of current, emphasizing that potential difference is not always constant.
Ohm's Law
Ohm's Law is an essential principle in electrical circuits, providing a relationship between current, voltage, and resistance. It is fundamental for analyzing circuits and solving problems related to electrical behavior.
The law states that the current flowing through most materials is directly proportional to the voltage across it and inversely proportional to its resistance.
  • Expressed mathematically, Ohm's Law is \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
  • In our problem, internal resistance is a crucial aspect. It affects how potential difference and current behavior changes inside the battery. Therefore, influencing the overall voltage output across the terminals.
  • For batteries, Ohm's Law helps define how well they can deliver current under different conditions by factoring in internal resistance.
  • Understanding Ohm's Law is vital for solving any circuit problem as it connects the three primary electrical properties: voltage, current, and resistance, making it easier to derive unknowns from known quantities.

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Most popular questions from this chapter

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours (A dot h). A 50-A dot h battery can supply a current of 50 A for 1.0 h, or 25 A for 2.0 h, and so on. (a) What total energy can be supplied by a 12-V, 60-A dot h battery if its internal resistance is negligible? (b) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 900 kg/m\(^3\).) (c) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully?

A lightning bolt strikes one end of a steel lightning rod, producing a 15,000-A current burst that lasts for 65 \(\mu\)s. The rod is 2.0 m long and 1.8 cm in diameter, and its other end is connected to the ground by 35 m of 8.0-mm-diameter copper wire. (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

A typical small flashlight contains two batteries, each having an emf of 1.5 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 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.)

A cylindrical copper cable 1.50 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 90.0 W? (b) What is the electric field inside the cable under these conditions?

In another experiment, a piece of the web is suspended so that it can move freely. When either a positively charged object or a negatively charged object is brought near the web, the thread is observed to move toward the charged object. What is the best interpretation of this observation? The web is (a) a negatively charged conductor; (b) a positively charged conductor; (c) either a positively or negatively charged conductor; (d) an electrically neutral conductor.
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