Textbook Question
In the circuit shown in Fig. E26.23, ammeter A1 reads 10.0 A and the batteries have no appreciable internal resistance. What is the resistance of R?
Ch 26: Direct-Current Circuits
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors41
- Ch 02: Motion Along a Straight Line67
- Ch 03: Motion in Two or Three Dimensions47
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws59
- Ch 06: Work & Kinetic Energy14
- Ch 07: Potential Energy & Conservation8
- Ch 08: Momentum, Impulse, and Collisions18
- Ch 09: Rotation of Rigid Bodies42
- Ch 10: Dynamics of Rotational Motion48
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics31
- Ch 13: Gravitation35
- Ch 14: Periodic Motion32
- Ch 15: Mechanical Waves25
- Ch 16: Sound & Hearing32
- Ch 17: Temperature and Heat36
- Ch 18: Thermal Properties of Matter38
- Ch 19: The First Law of Thermodynamics33
- Ch 20: The Second Law of Thermodynamics22
- Ch 21: Electric Charge and Electric Field36
- Ch 22: Gauss' Law24
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF26
- Ch 26: Direct-Current Circuits30
- Ch 27: Magnetic Field and Magnetic Forces18
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction28
- Ch 30: Inductance17
- Ch 31: Alternating Current21
- Ch 32: Electromagnetic Waves1
- Ch 33: The Nature and Propagation of Light20
- Ch 34: Geometric Optics34
- Ch 35: Interference19
- Ch 36: Diffraction25
- Ch 37: Special Relativity20
- Ch 38: Photons: Light Waves Behaving as Particles15
- Ch 39: Particles Behaving as Waves26
- Ch 40: Quantum Mechanics I: Wave Functions21
- Ch 41: Quantum Mechanics II: Atomic Structure25
- Ch 42: Molecules and Condensed Matter18
- Ch 43: Nuclear Physics16
- Ch 44: Particle Physics and Cosmology23
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 26, Problem 24a
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
In the circuit shown in Fig. E26.27 find the current in the 3.00 Ω resistor.
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
In the circuit shown in Fig. E26.27 find the unknown emfs and .