Textbook Question
A triangular array of resistors is shown in Fig. E26.5. What current will this array draw from a 35.0-V battery having negligible internal resistance if we connect it across (b) bc?
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Ch 26: Direct-Current Circuits
Chapter 26, Problem 26
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) If the power rating of a 15-kΩ resistor is 5.0 W, what is the maximum allowable potential difference across the termi-nals of the resistor?
Verified step by step guidance1
Identify the given values: resistance (R) = 15 kΩ = 15000 Ω, and power rating (P) = 5.0 W.
Recall the power formula related to resistance and voltage: P = V^2 / R, where P is the power, V is the voltage across the resistor, and R is the resistance.
Rearrange the formula to solve for V (voltage): V = sqrt(P * R).
Substitute the given values into the rearranged formula: V = sqrt(5.0 W * 15000 Ω).
Calculate the square root to find the maximum allowable potential difference across the resistor.
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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, where V is the voltage, I is the current, and R is the resistance. Understanding this law is essential for calculating the voltage across a resistor when given its resistance and the power it dissipates.
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Resistance and Ohm's Law
Power in Electrical Circuits
The power (P) dissipated by a resistor in an electrical circuit can be calculated using the formula P = IV, where I is the current through the resistor and V is the voltage across it. Additionally, using Ohm's Law, power can also be expressed as P = I²R or P = V²/R. This concept is crucial for determining the maximum voltage that can be applied to a resistor without exceeding its power rating.
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Power Rating of Resistors
The power rating of a resistor indicates the maximum amount of power it can safely dissipate without overheating. Exceeding this rating can lead to increased temperature, potentially damaging the resistor. In this context, knowing the power rating allows for the calculation of the maximum allowable voltage across the resistor, ensuring safe operation within its specified limits.
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Related Practice
Textbook Question
A triangular array of resistors is shown in Fig. E26.5. What current will this array draw from a 35.0-V battery having negligible internal resistance if we connect it across (c) ac?
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Textbook Question
A triangular array of resistors is shown in Fig. E26.5. (d) If the battery has an internal resistance of 3.00Ω, what current will the array draw if the battery is connected across bc?
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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. (b) 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. (c) 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 (a) the current through each bulb.
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