Ch 25: Current, Resistance, and EMF

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 25: Current, Resistance, and EMF

Problem 39a

Chapter 25, Problem 39a

Consider the circuit of Fig. E25.30. What is the total rate at which electrical energy is dissipated in the 5.0 Ω and 9.0 Ω resistors?

Verified step by step guidance

First, identify the resistors in the circuit that are relevant to the problem. The problem asks for the power dissipated in the 5.0-Ω and 9.0-Ω resistors. However, the image does not show these resistors directly. We need to analyze the circuit to find the equivalent resistance and current through the circuit.

Calculate the total voltage in the circuit. The circuit has two batteries: one with 12.0 V and another with 7.50 V. Since they are in series, the total voltage is the sum of these two voltages. Use the formula:

V = V

₁+V₂.

Determine the total resistance in the circuit. The resistors are in series, so the total resistance is the sum of all resistors: 3.2 Ω, 1.0 Ω, 7.0 Ω, and 4.0 Ω. Use the formula:

R

_total=R₁+R₂+R₃+R₄.

Calculate the current flowing through the circuit using Ohm's Law, which states that

I = \frac{V}{R}

. Substitute the total voltage and total resistance into this formula to find the current.

Finally, calculate the power dissipated in each resistor using the formula for power:

P = I^{2} R

. Apply this formula to each resistor individually to find the power dissipated in the 5.0-Ω and 9.0-Ω resistors, if they were present. Since the image does not show these resistors, focus on the resistors given in the image.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ohm's Law

Ohm's Law is fundamental in circuit analysis, stating that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. It is expressed as V = IR, and is essential for calculating current in each resistor to determine power dissipation.

Power Dissipation in Resistors

Power dissipation in resistors is the rate at which electrical energy is converted into heat, calculated using the formula P = I^2R, where P is power, I is current, and R is resistance. Understanding this concept is crucial for determining how much energy is lost in the 5.0-Ω and 9.0-Ω resistors in the circuit.

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law states that the sum of the electrical potential differences (voltage) around any closed network is zero. This principle helps in analyzing complex circuits by allowing the calculation of unknown voltages and currents, ensuring that the total voltage supplied equals the total voltage drop across the circuit components.

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Kirchhoff's Junction Rule

Related Practice

Textbook Question

The circuit shown in Fig. E $25.33$ contains two batteries, each with an emf and an internal resistance, and two resistors. Find the current in the circuit (magnitude and direction).

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Textbook Question

When a resistor with resistance $R$ is connected to a $1.50$ -V flashlight battery, the resistor consumes $0.0625$ W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) What power does the resistor consume if it is connected to a $12.6$ -V car battery? Assume that $R$ remains constant when the power consumption changes.

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Textbook Question

The circuit shown in Fig. E $25.33$ contains two batteries, each with an emf and an internal resistance, and two resistors. Find the terminal voltage $V_{a b}$ of the $16.0$ -V battery.

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Textbook Question

Consider the circuit of Fig. E25.30. What is the power output of the 16.0 V battery?

Textbook Question

Consider the circuit of Fig. E25.30 Show that the power output of the 16.0 V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.

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Textbook Question

Consider the circuit of Fig. E25.30. At what rate is electrical energy being converted to other forms in the 8.0 V battery?

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