Textbook Question
The batteries shown in the circuit in Fig. E26.24 have negligibly small internal resistances. Find the current through the 30.0-Ω resistor.
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 23a
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?
Verified step by step guidance1
First, identify the components in the circuit. You have an ammeter A1 reading 10.0 A, a resistor R, and batteries with no internal resistance. This means the voltage across the batteries is fully applied to the circuit.
Use Ohm's Law, which states that \( V = I \times R \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. Since the ammeter reads 10.0 A, this is the current \( I \) flowing through the resistor R.
Determine the total voltage \( V \) supplied by the batteries. If the problem provides the voltage value, use it directly. If not, you may need additional information from the circuit diagram or problem context.
Substitute the known values into Ohm's Law to solve for the resistance \( R \). Rearrange the formula to \( R = \frac{V}{I} \). Plug in the current \( I = 10.0 \) A and the voltage \( V \) to find \( R \).
Verify your solution by checking the units and ensuring that the calculated resistance makes sense in the context of the circuit. If additional components or information are provided, consider their impact on the circuit and the resistance calculation.
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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 that states the relationship between voltage, current, and resistance in a circuit: V = IR, where V is voltage, I is current, and R is resistance. This law is essential for calculating the resistance in a circuit when the current and voltage are known.
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Resistance and Ohm's Law
Series and Parallel Circuits
Understanding series and parallel circuits is crucial for analyzing how components like resistors affect the overall circuit. In a series circuit, resistances add up, while in a parallel circuit, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances. This concept helps determine how the current flows and how resistances are calculated.
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Kirchhoff's Circuit Laws
Kirchhoff's Circuit Laws include the Current Law (KCL) and the Voltage Law (KVL). KCL states that the total current entering a junction equals the total current leaving, while KVL states that the sum of the electrical potential differences around any closed network is zero. These laws are vital for analyzing complex circuits and determining unknown values like resistance.
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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
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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