17.1 电电路的能源转让
Section outline
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Part of the , an electrical transmission sub-station receives extremely high current levels, then passes the electrical energy on to as many as 200,000 homes. Approximately 5000 megawatt-hours of passes through this particular substation each year.
::部分电力传输分站目前的电量极高,然后将电能输送到多达200,000个家庭,每年通过该分站的电流量约为5,000兆瓦小时。Energy Transfer in Electric Circuits
::电电路的能源转让Electric power is the energy per unit time converted by an electric into another form of energy. We already know that power through a circuit is equal to the multiplied by the current in a circuit: @$\begin{align*}P = VI\end{align*}@$ . It is possible to determine the power dissipated in a single resistor if we combine this expression with Ohm’s Law , @$\begin{align*}V = IR\end{align*}@$ . This becomes particularly useful in with more than one resistor, to determine the power dissipated in each one. Combining these two equations, we get an expression for electric power that involves only the current and in a circuit.
::电能是用电转换成另一种能源形式的单位时间的能量。 我们已经知道电路电能等于电流乘以电流的倍数, 在电路中是 : @ $\ begin{ align}P = VI\ end{ align}$。 如果我们将这个表达式与 Ohm 的法律相结合, @ $\ begin{ align}V = IR\ end{ allign}$ 。 这在多个抵抗器中变得特别有用, 以决定每条电流的耗电量。 结合这两个方程式, 我们可以得到只涉及电流和电路的电力表达式 。@$$\begin{align*}P = I^2R\end{align*}@$$
::@ $\ begin{ align} p = I_ 2R\ end{ aliign} $The power dissipated in a resistor is proportional to the square of the current that passes through it and to its resistance.
::抵抗者体内的能量消散 与流经的海流平方和抵抗力成正比Electrical energy itself can be expressed as the electrical power multiplied by time:
::电能本身可以用电力乘以时间表示:@$\begin{align*}E = Pt\end{align*}@$
::@$\ begin{ aliign} E = Pt\ end{ aliign} $ @ $We can incorporate this equation to obtain an equation for electrical energy based on current, resistance, and time. The electrical energy across a resistor is determined to be the current squared multiplied by the resistance and the time.
::我们可以纳入这个等式,以获得基于电流、阻力和时间的电能等式。 阻力器的电能被确定为电流乘以阻力和时间。@$\begin{align*}E = I^2Rt\end{align*}@$
::@ $\ begin{ aliign} E = I_ 2Rt\ end{ aliign} $This equation holds true in ideal situations. However, devices used to convert electrical energy into other are never 100% efficient. An is used to convert electrical energy into , but some of the electrical energy in this process is lost to . When a lamp converts electrical energy into light energy, some electrical energy is lost to thermal energy.
::这个等式在理想情况下是正确的。 然而, 用于将电能转换成其他能源的装置从来就没有百分之百的效率。 电能转换成电能, 但在此过程中部分电能丢失到 。 当灯把电能转换成光能时, 一些电能会丢失到热能中 。Example 1
::例1A heater has a resistance of 25.0 Ω and operates on 120.0 V.
::一个加热器的抗药性为25.0 ,在120.0 V上运行。-
How much current is supplied to the resistance?
::抵抗军有多少电流? -
How many
joules
of energy is provided by the heater in 10.0 s?
::热器在10秒内能提供多少焦耳?
-
@$\begin{align*}I=\frac{V}{R}=\frac{120.0 \ V}{25.0 \ \Omega}=4.8 \ A\end{align*}@$
::@ $\ begin{ legign} I @ frac{ VRfrac} 120.0\ V25. 0\\\\ Omega} @ 4. 8\ A\end{ lair$} @ $ -
@$\begin{align*}E = I^2Rt = (4.8 \ A)^2(25.0 \ \Omega)(10.0 \ s) = 5760 \ \text{joules}\end{align*}@$
::@ $\ begin{ legign} E = I# 2Rt = (4. 8 \ A) 2: 25. 0 \ \ \ 欧米茄 (1.0 0\ s) = 5760\ text{ joules\ end{ align$} = 5760\ \ text{ joules\ end{ end{ align$}
Think again about the power grid. When electricity is transmitted over long distances, some amount of energy is lost in overcoming the resistance in the transmission lines. We know the equation for the power dissipated is given by @$\begin{align*}P = I^2R\end{align*}@$ . The energy loss can be minimized by choosing the material with the least resistance for power lines, but changing the current also has significant effects. Consider a reduction of the current by a power of ten:
::再想想电网。当电力通过长距离传输时,在克服输电线阻力方面损失了一定的能量。 我们知道耗电的方程式是由 @$\ begin{ alignQP = I# 2R\ end{ ALign}$提供的。 选择电线阻力最小的材料可以将能量损失降到最低, 但改变电流也会产生重大影响。 想象一下, 电流减少十倍 :How much power is dissipated when a current of 10.0 A passes through a power line whose resistance is 1.00 Ω? @$\begin{align*}P = I^2R = (10.0 \ A)^2(1.00 \ \Omega) = 100. \ \text{Watts}\end{align*}@$
::当一个10.0A的电流通过一个抗力为1.00的电线时,能耗多少? @$\ begin{align}P = I#2R = (1.0 \ A)\2\ 1. 00\\\ Omega) = 100.\ text{Watts}end{align}= 100.\ text{Watts}end{align}How much power is dissipated when a current of 1.00 A passes through a power line whose resistance is 1.00 Ω? @$\begin{align*}P = I^2R = (1.00 \ A)^2(1.00 \ \Omega) = 1.00 \ \text{Watts}\end{align*}@$
::当1.00 A 的电流通过一个阻力为 1. 00 的电线时, 电流的功率会消失多少? @ $\ begin{ align\\ p = I# 2R = (1. 00\ A)\\ 2( 1. 00\\ \ Omega) = 1. 00\ text{ Watts}end{ ALign} $The power loss is reduced tremendously by reducing the magnitude of the current through the resistance. Power companies must transmit the same amount of energy over the power lines but keep the power loss minimal. They do this by reducing the current. From the equation @$\begin{align*}P = VI\end{align*}@$ , we know that the voltage must be increased to keep the same power level.
::通过阻力降低电流的强度,电力损失会大大降低。 电力公司必须在电线上传输同样数量的能源, 但将电力损失控制在最小范围内。 他们通过减少电流来做到这一点。 从公式@$\ begin{ align}P = VI\ end{ ALign}$, 我们知道必须增加电压以保持同样的电量水平 。The Kilowatt-Hour
::基洛瓦特时Even though the companies that supply electrical energy are often called “power” companies, they are actually selling energy. Your electricity bill is based on energy, not power. The amount of energy provided by electric current can be calculated by multiplying the watts (J/s) by seconds to yield joules. The joule, however, is a very small unit of energy and using the joule to state the amount of energy used by a household would require a very large number. For that reason, electric companies measure their energy sales in a large number of joules called a kilowatt hour (kWh). A kilowatt hour is exactly as it sounds - the number of kilowatts (1,000 W) transferred per hour.
::尽管电力供应公司通常被称为“电力”公司,但它们实际上是在销售能源。你的电力账单是基于能源,而不是电力。电流提供的能源量可以通过将瓦特乘以秒来计算,以产生焦耳。然而,焦耳是一个很小的能源单位,用焦耳来说明家庭使用的能源量需要非常大的数量。因此,电力公司用大量焦耳来衡量其能源销售量,称为千瓦小时。千瓦小时恰好是每小时传输千瓦小时的数量。@$\begin{align*}1.00 \ \text{kilowatt hour} = (1000 \ J/s)(3600 \ s) = 3.6 \times 10^6 \ J\end{align*}@$
::@ $\ begin{ align\\ 1. 00\\ text{ kilowatt hour} = (1000\ J/s)( 3600\ s) = 3. 6\ times 10\ 6\ J\ end{ leign} = 3.6\ times 10\ 6\ J\ end{ end{ leign} = (1000\ J/s)( 3600\ s) = 3.6\ times 10\ 6\ \ end{ leign} $Example 2
::例2A color television uses about 2.0 A when operated on 120 V.
::一个彩色电视在120伏下操作时使用约2.0A。-
How much power does the set use?
::装配装置能用多少电源? -
If the TV is operated for 8.00 hours per day, how much energy in kWh does it use per day?
::如果电视每天运行8小时,每天用多少千瓦时的能量? -
At $0.15 per kWh, what does it cost to run the TV for 30 days?
::每千瓦时0.15美元, 30天的电视运行成本是多少?
-
@$\begin{align*}P = VI = (120 \ V)(2.0 \ A) = 240 \ W\end{align*}@$
::@$\ begin{ legign} = VI = (120\ V)(2. 0\ A) = 240\ W\ end{ legign} = VI = (120\ V)(2. 0\ A) = 240\ W\ end{ legign} $ -
@$\begin{align*}E=\frac{(240 \ J/s)(8 \ h)(3600 \ s/h)}{3.6 \times 10^6 \ J/kWh}=1.92 \ kWh\end{align*}@$
::@$\ begin{ legin}Efrac{ (240\ J/s) 8\ h (3600\ s/h)\\ 3. 6\ times 10\ 6\ J/kWh\\ 1. 92\ kWh\ end{ align} $ -
@$\begin{align*}\text{Cost} = (1.92 \ kWh)(30)(\$0.15) = \$8.64\end{align*}@$
::@ $\ begin{ align} text{ Cost} = (1. 92\ kWh)( 30) (\ $0. 15) =\ $8. 64\ end{ end{ leign} = (1. 92\ kWh)( 30) (\ $. 0. 15) =\ $8. 64\ end{ end{ lairign} $
Launch the Dollhouse simulation below and click Show Power to observe the energy consumed by the dollhouse. Try to adjust the Potential slider and Device sliders to maximize the total power consumed by the dollhouse:
::在下面启动 Dollhouse 模拟, 并单击 Show Power 以观察玩偶屋消耗的能量。 试着调整潜在滑动器和设备滑动器, 以尽量扩大玩偶屋消耗的总电源 :Further Reading
::继续阅读-
Electric
Power
and
Electrical
Energy
Use
::电力和电力能源使用 -
Efficiency
::效率效率效率
Summary
::摘要-
Electric power is the energy per unit time converted by an electric circuit into another form of energy.
::电力是电路转换成另一种能源的每单位时间的能量。 -
The formula for electric power is
@$\begin{align*}P = I^2R\end{align*}@$
.
::电源公式是 @$\ begin{ align} p = I2R\ end{ aliign} $ 。 -
The electric energy transferred to a resistor in a time period is equal to the electric power multiplied by time,
@$\begin{align*}E = Pt\end{align*}@$
, and can also be calculated using
@$\begin{align*}E = I^2Rt\end{align*}@$
.
::在一段时间内传输到抵抗器的电能等于乘以时间的电能,@$\ begin{ align}_E = Pt\end{ align}_$,也可以使用@$\ begin{ align}_E = I_2Rt\end{align}$计算。 -
Electric companies measure their energy sales in a large number of joules called a kilowatt hour (kWh) which is equivalent to
@$\begin{align*}3.6 \times 10^6 \ J\end{align*}@$
.
::电力公司衡量其能源销售量的约尔斯数量庞大,称为千瓦时,相当于@$\ begin{align}3.6\ times 106\ J\end{align}$。
Review
::回顾-
A 2-way light bulb for a 110. V lamp has filament that uses power at a rate of 50.0 W and another filament that uses power at a rate of 100. W. Find the resistance of these two filaments.
::一台110V型灯的双向灯泡有丝质,用电率为50.0瓦,另一台用电率为100瓦。 找出这两个丝质的阻力。 -
Find the power dissipation of a 1.5 A lamp operating on a 12 V battery.
::找到一个1.5A灯的电源耗竭 在一个12V电池上操作 -
A high voltage
@$\begin{align*}(4.0 \times 10^5 \ V)\end{align*}@$
power transmission line delivers electrical energy from a generating station to a substation at a rate of
@$\begin{align*}1.5 \times 10^9 \ W\end{align*}@$
. What is the current in the lines?
::高电压@ $\ begin{ begin{ align}} (4. 0\ times 10_ 5\ V)\ end{ align}$ power 输电线以@$\ begin{ aliign} 1.5\ times 10_ 9\ W\end{ align}$的速率将电力从发电站输送到分站。 该行的当前情况是什么 ? -
A toaster oven indicates that it operates at 1500 W on a 110 V circuit. What is the resistance of the oven?
::烤面包机烤箱表明它在110V电路1500W上操作。烤箱的抗力是什么?
Explore More
::探索更多Use this resource to answer the questions that follow.
::使用此资源回答下面的问题 。-
What is the definition of electrical power?
::电力的定义是什么? -
What happens to the electrical energy that is not converted into work?
::电能不转化为工作又会怎样呢?
-
How much current is supplied to the resistance?
