Textbook Question
The 5.00 V battery in Fig. E26.28 is removed from the circuit and replaced by a 15.00 V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure. Find the current in each branch.
Ch 26: Direct-Current Circuits
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors37
- Ch 02: Motion Along a Straight Line57
- Ch 03: Motion in Two or Three Dimensions45
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws54
- Ch 06: Work & Kinetic Energy11
- Ch 07: Potential Energy & Conservation6
- Ch 08: Momentum, Impulse, and Collisions15
- Ch 09: Rotation of Rigid Bodies37
- Ch 10: Dynamics of Rotational Motion44
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics27
- Ch 13: Gravitation34
- Ch 14: Periodic Motion26
- Ch 15: Mechanical Waves22
- Ch 16: Sound & Hearing28
- Ch 17: Temperature and Heat28
- Ch 18: Thermal Properties of Matter35
- Ch 19: The First Law of Thermodynamics29
- Ch 20: The Second Law of Thermodynamics18
- Ch 21: Electric Charge and Electric Field28
- Ch 22: Gauss' Law21
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF24
- Ch 26: Direct-Current Circuits29
- Ch 27: Magnetic Field and Magnetic Forces16
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction22
- Ch 30: Inductance14
- Ch 31: Alternating Current20
- Ch 33: The Nature and Propagation of Light17
- Ch 34: Geometric Optics29
- Ch 35: Interference13
- Ch 36: Diffraction19
- Ch 37: Special Relativity19
- Ch 38: Photons: Light Waves Behaving as Particles12
- Ch 39: Particles Behaving as Waves25
- Ch 40: Quantum Mechanics I: Wave Functions20
- Ch 41: Quantum Mechanics II: Atomic Structure21
- Ch 42: Molecules and Condensed Matter17
- Ch 43: Nuclear Physics13
- Ch 44: Particle Physics and Cosmology17
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 26, Problem 32b
In the circuit shown in Fig. E26.34, the 6.0 Ω resistor is consuming energy at a rate of 24 J/s when the current through it flows as shown. What are the polarity and emf ε of the unknown battery, assuming it has negligible internal resistance?
Verified step by step guidance1
First, understand that the power consumed by the resistor is given by the formula: , where is the power, is the current, and is the resistance.
Given that the power is 24 J/s and the resistance is 6.0 Ω, use the formula to solve for the current : .
Once you have the current, apply Kirchhoff's loop rule to the circuit. This rule states that the sum of the potential differences (voltage) around any closed loop in a circuit must be zero.
Identify the direction of the current flow and the polarity of the known battery in the circuit. Use the loop rule to write an equation that includes the emf of the unknown battery, the voltage drop across the resistor, and any other known voltages.
Solve the equation for the unknown emf . The sign of will indicate the polarity of the unknown battery. If is positive, the polarity is in the direction of the assumed current flow; if negative, it is opposite.
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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 states that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance. It is expressed as V = IR, where V is voltage, I is current, and R is resistance. This law is essential for calculating the current and voltage in the circuit.
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03:07
Resistance and Ohm's Law
Power in Electrical Circuits
Power in electrical circuits is the rate at which energy is consumed or converted. It is calculated using the formula P = IV, where P is power, I is current, and V is voltage. Alternatively, it can be expressed as P = I^2R or P = V^2/R, depending on known values. Understanding power consumption helps determine the current flowing through the resistor.
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Electromotive Force (EMF)
Electromotive force (EMF) is the energy provided by a battery or power source per coulomb of charge. It is the potential difference across the terminals of a battery when no current is flowing. EMF is crucial for determining the voltage supplied by the unknown battery in the circuit, influencing the current and power distribution.
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Related Practice
Textbook Question
The heating element of an electric dryer is rated at 4.1 kW when connected to a 240 V line. What is the resistance of the dryer's heating element at its operating temperature?
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Textbook Question
A 1500 W electric heater is plugged into the outlet of a 120 V circuit that has a 20 A circuit breaker. You plug an electric hair dryer into the same outlet. The hair dryer has power settings of 600 W, 900 W, 1200 W, and 1500 W. You start with the hair dryer on the 600 W setting and increase the power setting until the circuit breaker trips. What power setting caused the breaker to trip?
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Textbook Question
An emf source with ε = 120V, a resistor with R = 80.0Ω, and a capacitor with C = 4.00 μF are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?
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Textbook Question
In the circuit shown in Fig. E26.31 the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 15.0 V. What will the ammeter read when the switch is closed?
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Textbook Question
In the circuit shown in Fig. E26.33 all meters are idealized and the batteries have no appreciable internal resistance. Find the reading of the voltmeter with the switch S open. Which point is at a higher potential: a or b?