Ch. 25 - Electric Current and Resistance

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 25 - Electric Current and Resistance

Problem 41

Chapter 24, Problem 41

(II) A power station delivers 750 kW of power at 12,000 V to a factory through wires with total resistance 3.0 Ω. How much less power is wasted if the electricity is delivered at 50,000 V rather than 12,000 V?

Verified step by step guidance

Step 1: Calculate the current (I) in the circuit when the power is delivered at 12,000 V. Use the formula for power:

P = V I

, where P is the power (750 kW) and V is the voltage (12,000 V). Rearrange to find

I = P / V

Step 2: Calculate the power wasted in the wires when the voltage is 12,000 V. Use the formula for power loss due to resistance:

P_{loss} = I^{2} R

, where I is the current calculated in Step 1 and R is the total resistance of the wires (3.0 Ω).

Step 3: Repeat Step 1 to calculate the current (I) when the power is delivered at 50,000 V. Use the same formula:

I = P / V

, but substitute 50,000 V for the voltage.

Step 4: Calculate the power wasted in the wires when the voltage is 50,000 V. Again, use the formula

P_{loss} = I^{2} R

, but use the current calculated in Step 3.

Step 5: Find the difference in power wasted between the two cases. Subtract the power loss at 50,000 V from the power loss at 12,000 V to determine how much less power is wasted when the electricity is delivered at the higher voltage.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power and Voltage Relationship

The relationship between power (P), voltage (V), and current (I) is defined by the equation P = VI. This means that for a given power output, increasing the voltage allows for a decrease in current, which is crucial for minimizing energy losses in transmission lines.

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, expressed as V = IR. This principle helps in understanding how resistance affects power loss in electrical systems.

Power Loss in Transmission Lines

Power loss in transmission lines is primarily due to the resistance of the wires, calculated using the formula P_loss = I²R. As current increases, the power lost as heat in the wires also increases, making high-voltage transmission more efficient by reducing current and thus minimizing losses.

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Related Practice

Textbook Question

You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 120 V (assume its resistance does not change), how bright will it be relative to 75-W 120-V bulbs? [Hint: Assume roughly that brightness is proportional to power consumed.]

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Textbook Question

A flashlight uses two AA 1.5-V batteries connected in series to provide 3.0 V across the bulb, as in Fig. 25–4b. The bulb draws 135 mA when turned on.

(a) Calculate the resistance of the bulb and the power dissipated.

(b) By what factor would the power increase if four AA batteries in series (total 6.0 V) were used with the same bulb? (Neglect heating effects of the filament.) Why shouldn’t you try this?

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Textbook Question

At \$0.12/kWh, what does it cost to leave a 25-W porch light on day and night for a year?

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Textbook Question

Calculate the peak current in a 2.5-k Ω resistor connected to a 220-V rms ac source.

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Textbook Question

Determine the maximum current passing through a 2.7-hp pump connected to a 240-V_rms ac power source.

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Textbook Question

A heater coil connected to a 240-Vᵣₘₛ ac line has a resistance of 32Ω. What are the maximum and minimum values of the instantaneous power?

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