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Ch 28: Fundamentals of Circuits
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 28: Fundamentals of CircuitsProblema 50
Capitolo 28, Problema 50

A lightbulb is in series with a 2.0 Ω resistor. The lightbulb dissipates 10 W when this series circuit is connected to a 9.0 V battery. What is the current through the lightbulb? There are two possible answers; give both of them.

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Step 1: Start by recalling the formula for power dissipation in terms of current and resistance: \( P = I^2 R \). Here, \( P \) is the power dissipated, \( I \) is the current, and \( R \) is the resistance. The lightbulb dissipates 10 W, so substitute \( P = 10 \) W into the equation.
Step 2: Rearrange the formula \( P = I^2 R \) to solve for \( I \): \( I = \sqrt{\frac{P}{R}} \). To proceed, you need the resistance of the lightbulb, which can be determined using the total voltage and the series circuit properties.
Step 3: Use Ohm's Law \( V = IR \) and the formula for total resistance in a series circuit: \( R_{total} = R_{resistor} + R_{lightbulb} \). The total voltage is 9.0 V, and the resistor has a resistance of 2.0 Ω. Let \( R_{lightbulb} \) represent the unknown resistance of the lightbulb.
Step 4: The total power dissipated in the circuit is shared between the resistor and the lightbulb. Use the relationship \( P = IV \) to express the current in terms of the total voltage and resistance. Substitute \( R_{total} \) into the equations to find \( R_{lightbulb} \).
Step 5: Once \( R_{lightbulb} \) is determined, substitute it back into \( I = \sqrt{\frac{P}{R}} \) to calculate the current through the lightbulb. Since there are two possible values for \( R_{lightbulb} \) (due to the quadratic nature of the equations), you will find two possible values for \( I \).

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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. It is mathematically expressed as V = I * R. This principle is fundamental in analyzing electrical circuits, as it helps determine the relationship between voltage, current, and resistance.
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Resistance and Ohm's Law

Power in Electrical Circuits

The power (P) dissipated in an electrical component, such as a lightbulb or resistor, is given by the formula P = V * I, where V is the voltage across the component and I is the current flowing through it. In this context, knowing the power dissipated by the lightbulb allows us to calculate the current using the voltage supplied by the battery, which is essential for solving the problem.
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Power in Circuits

Series Circuits

In a series circuit, components are connected end-to-end, so the same current flows through each component. The total resistance in a series circuit is the sum of the individual resistances. This concept is crucial for understanding how the lightbulb and resistor interact in the circuit, as the total resistance affects the current and voltage distribution across each component.
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LRC Circuits in Series
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