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Ch 26: Direct-Current Circuits
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 26: Direct-Current CircuitsProblem 21a
Chapter 26, Problem 21a

Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. If the two light bulbs are connected in series across a 120-V line, find the current through each bulb.

Verified step by step guidance
1
First, understand that when resistors (or light bulbs in this case) are connected in series, the total resistance is the sum of the individual resistances. So, calculate the total resistance \( R_{total} \) using the formula: \( R_{total} = R_1 + R_2 \), where \( R_1 = 400 \Omega \) and \( R_2 = 800 \Omega \).
Next, apply Ohm's Law to find the current through the series circuit. Ohm's Law states that \( V = I \times R \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. Rearrange the formula to solve for current \( I \): \( I = \frac{V}{R_{total}} \). Use the total resistance calculated in step 1 and the given voltage \( V = 120 \text{ V} \).
Since the bulbs are in series, the current through each bulb is the same. Therefore, the current calculated in step 2 is the current through each bulb.
To verify your understanding, consider the implications of series circuits: the same current flows through each component, and the voltage across the entire circuit is the sum of the voltages across each component. This confirms that the current through each bulb is indeed the same.
Finally, reflect on the concept of series circuits and how the total resistance affects the current. In series circuits, increasing the resistance decreases the current, given a constant voltage. This is a fundamental principle that can be applied to various problems involving series circuits.

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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 I = V/R, where I is the current, V is the voltage, and R is the resistance. This principle is essential for calculating the current in circuits.
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Series Circuit

In a series circuit, components are connected end-to-end, forming a single path for current flow. The total resistance in a series circuit is the sum of individual resistances, and the same current flows through each component. Understanding series circuits is crucial for determining how voltage and current are distributed across components.
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Equivalent Resistance

Equivalent resistance in a series circuit is the sum of all resistances. For resistors in series, the formula is R_eq = R1 + R2 + ... + Rn. This concept helps in simplifying complex circuits into a single resistance value, making it easier to apply Ohm's Law to find the current through the circuit.
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Related Practice
Textbook Question

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. If the power rating of a 15 kΩ resistor is 5.0 W, what is the maximum allowable potential difference across the terminals of the resistor?

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Textbook Question

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. A 9.0 kΩ resistor is to be connected across a 120 V potential difference. What power rating is required?

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Textbook Question

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. A 100.0 Ω and a 150.0 Ω resistor, both rated at 2.00 W, are connected in series across a variable potential difference. What is the greatest this potential difference can be without overheating either resistor, and what is the rate of heat generated in each resistor under these conditions?

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Textbook Question

Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. If the two light bulbs are connected in series across a 120 V line, find the power dissipated in each bulb.

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Textbook Question

Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. If the two light bulbs are connected in series across a 120 V line, find the total power dissipated in both bulbs.

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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 current through each bulb.

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