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Ch 26: Direct-Current Circuits
Chapter 26, Problem 26

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 (f) the total power dissipated in both bulbs.

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1
Calculate the total resistance of the two bulbs when connected in parallel using the formula: \( R_{\text{total}} = \left( \frac{1}{R_1} + \frac{1}{R_2} \right)^{-1} \), where \( R_1 = 400 \Omega \) and \( R_2 = 800 \Omega \).
Use the total resistance calculated in step 1 to find the total current flowing through the circuit using Ohm's Law: \( I = \frac{V}{R_{\text{total}}} \), where \( V = 120 \, \text{V} \).
Calculate the power dissipated by each bulb separately using the formula: \( P = I^2 \times R \), where \( I \) is the current calculated in step 2 and \( R \) is the resistance of each bulb.
Add the power dissipated by each bulb to find the total power dissipated in the circuit.
Verify the calculations by checking the units and ensuring that the total power dissipated makes sense given the resistances and the voltage.

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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 (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. This relationship is expressed as V = IR, which is fundamental for analyzing electrical circuits, including those with resistors in series or parallel.
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Resistors in Parallel

When resistors are connected in parallel, the total or equivalent resistance (R_total) can be calculated using the formula 1/R_total = 1/R1 + 1/R2 + ... + 1/Rn. This configuration allows for multiple paths for current to flow, resulting in a lower total resistance compared to individual resistances, which affects the overall current and power dissipation in the circuit.
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Power Dissipation

Power dissipation in an electrical circuit is the rate at which electrical energy is converted into heat or light, typically measured in watts (W). For resistors, power can be calculated using the formula P = IV, where I is the current through the resistor and V is the voltage across it. In parallel circuits, the total power is the sum of the power dissipated by each resistor.
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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. If the two light bulbs are connected in series across a 120-V line, find (c) 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 (d) the current through 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. The two light bulbs are now connected in parallel across the 120-V line. Find (e) 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. 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. (g) 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. (h) In which situation is there a greater total light output from both bulbs combined?
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Textbook Question
In the circuit shown in Fig. E26.31 the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 15.0 V.

(b) What will the ammeter read when the switch is closed?
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