Textbook Question
In the circuit shown in Fig. E26.27 find the current in the 3.00 Ω resistor.
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 22a
The batteries shown in the circuit in Fig. E26.24 have negligibly small internal resistances. Find the current through the 30.0-Ω resistor.
Verified step by step guidance1
Identify the components in the circuit: two batteries (E1 and E2) and two resistors (R1 and R2). The resistors are labeled with their resistance values, R1 = 30.0 Ω and R2 = unknown.
Apply Kirchhoff's loop rule to the circuit. This rule states that the sum of the electromotive forces (emf) in any closed loop is equal to the sum of the potential drops (voltage) across the resistors in that loop.
Write the equation for the loop using Kirchhoff's rule: E1 - I * R1 - E2 - I * R2 = 0, where I is the current through the resistors.
Rearrange the equation to solve for the current I: I = (E1 - E2) / (R1 + R2).
Substitute the known values into the equation. You have R1 = 30.0 Ω, and you need to know the values of E1, E2, and R2 to find the current I through the 30.0 Ω resistor.
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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 a fundamental principle in electronics and electrical engineering, stating 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 mathematically expressed as I = V/R, where I is the current, V is the voltage, and R is the resistance. This law is crucial for calculating the current in circuits.
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Resistance and Ohm's Law
Kirchhoff's Loop Rule
Kirchhoff's Loop Rule, also known as Kirchhoff's Voltage Law, states that the sum of the electromotive forces (emf) and potential differences (voltage) around any closed loop in a circuit is zero. This principle is essential for analyzing complex circuits, as it allows for the calculation of unknown currents and voltages by setting up equations based on the conservation of energy within the loop.
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11:15Intro to Kirchhoff's Loop Rule
Series and Parallel Circuits
In a series circuit, components are connected end-to-end, so the same current flows through each component, and the total resistance is the sum of individual resistances. In a parallel circuit, components are connected across the same voltage source, and the total current is the sum of the currents through each component. Understanding these configurations is vital for analyzing how current and voltage distribute in a circuit.
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Related Practice
Textbook Question
Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are connected in series across a 120 V line. Afterwards, the two light bulbs are connected in parallel across the 120 V line. In each situation, which of the two bulbs glows the brightest?
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Textbook Question
Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are connected in series across a 120 V line. Afterwards, the two light bulbs are connected in parallel across the 120 V line. In which situation is there a greater total light output from both bulbs combined?
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Textbook Question
Find the emfs and in the circuit of Fig. E26.26, and find the potential difference of point relative to point .
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Textbook Question
Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are now connected in parallel across the 120 V line. Find the total power dissipated in both bulbs.
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Textbook Question
In the circuit shown in Fig. E26.27 find the unknown emfs and .