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Chapter 28: Problem 60

A car-battery charger runs off the \(120-\mathrm{V}\) rms AC power line and supplies \(10-\mathrm{A}\) DC at \(14 \mathrm{V}\). (a) If the charger is \(80 \%\) efficient in converting the line power to the DC power it supplies to the battery, how much current does it draw from the AC line? (b) If electricity costs \(9.5 \notin / \mathrm{kWh}\), how much does it cost to run the charger for 10 hours if the power factor is \(1 ?\)

Short Answer

Expert verified
a) \(1.46 A\) from the AC line. b) \(16.625\) cents to run the charger for 10 hours.

Step by step solution

01

Understand the Problem and Gather Data

A car battery charger is producing a DC output power of \(14 \mathrm{V}* 10 \mathrm{A} = 140 \mathrm{W}\). The input power is not known, but the efficiency is given as \(0.8\) or \(80 \% \) . The charger runs for \(10\) hours and electricity costs \(9.5 \notin / \mathrm{kWh}\).
02

Calculate the Input Power

Since the efficiency of the power conversion is \(80 \% \), it implies that \(80 \% \) of the AC input power is converted into DC output power. From the definition of efficiency, which is output power divided by the input power, rearrange the formula to solve for input power, yielding \[Input \ Power = \frac{Output \ Power}{Efficiency} = \frac{140W}{0.8} = 175 W\]
03

Calculate the Current Drawn

Now that the input power is known, calculate the current drawn from the AC line. The formula for power is \(Power = VI\), where \(V\) is Voltage and \(I\) is Current. Rearranging to solve for current yields \[I = \frac{Power}{Voltage} = \frac{175 W}{120 V} = 1.46 A\]. Remember that current is the ratio of power to voltage.
04

Calculate the Cost of Running the Charger

To determine the cost of running the charger for \(10\) hours, first convert power to kilowatts (since the cost is in cents/kWh), then multiply by the number of hours it runs, and lastly the cost per kilowatt hour. \[Cost = Power(kW) * Time(h) * Rate(\notin/kWh) = 175 W * \frac{1 kW}{1000 W} * 10 h * 9.5 \notin/kWh = 16.625 \notin \]. The charger would therefore cost \(16.625\) cents to run for \(10\) hours.

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