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 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 21g
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?
Verified step by step guidance1
Understand the concept of brightness in light bulbs: Brightness is related to the power consumed by the bulb, which can be calculated using the formula \( P = \frac{V^2}{R} \) for series and parallel circuits.
Calculate the equivalent resistance for the bulbs in series: When resistors are in series, the total resistance \( R_{total} \) is the sum of individual resistances. So, \( R_{total} = R_1 + R_2 = 400\,\Omega + 800\,\Omega = 1200\,\Omega \).
Determine the current through the bulbs in series: Use Ohm's Law \( I = \frac{V}{R_{total}} \) to find the current. With \( V = 120\,V \) and \( R_{total} = 1200\,\Omega \), calculate \( I \). Then, use \( P = I^2 \times R \) to find the power for each bulb.
Calculate the equivalent resistance for the bulbs in parallel: For parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances. \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \). Solve for \( R_{total} \).
Determine the power for each bulb in parallel: Since the voltage across each bulb is the same (120 V), use \( P = \frac{V^2}{R} \) for each bulb to find the power. Compare the power values to determine which bulb glows the brightest in parallel.
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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, represented by the formula V = IR, where V is voltage, I is current, and R is resistance. This principle helps determine the current flowing through each bulb in both series and parallel configurations.
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Resistance and Ohm's Law
Series and Parallel Circuits
In a series circuit, components are connected end-to-end, so the same current flows through each component, but the voltage is divided among them. In a parallel circuit, components are connected across the same voltage source, so each component receives the full voltage, but the current is divided based on resistance. Understanding these configurations is crucial for analyzing the brightness of the bulbs.
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Power Dissipation in Resistors
Power dissipation in a resistor is given by the formula P = IV, where P is power, I is current, and V is voltage. Alternatively, it can be expressed as P = I²R or P = V²/R. The brightness of a bulb is related to the power it dissipates, so calculating power for each bulb in both series and parallel setups helps determine which bulb glows brightest.
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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 now connected in parallel across the 120 V line. Find the power dissipated in each bulb.
1
views
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 current through each bulb.
1
views
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?
3
views
Textbook Question
Find the emfs and in the circuit of Fig. E26.26, and find the potential difference of point relative to point .
1
views
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.
1
views