# Tag Info

2

From the definition of work $$W = \int dx F$$ and $$P = \frac{dW}{dt}$$ you can see how we can arrive at $$P = F \frac{dx}{dt} = F v$$ (when considering only the absolute value). To understand it intuitively, imagine the case of a frictionless system in which the car can move at a certain speed without any opposing force. The power required to keep it at ...

1

Along the lines of what @CuriousOne said you can use Kirchhoff's laws to understand the current and voltage drop at each circuit element with any complexity you desire. Kirchhoff's laws deal with the fact of the total sum of currents and voltages in any circuit junction and loop, respectively, must equal zero. The second law essentially uses Ohm's law to ...

2

When you work "fairly hard", your body can produce about 200 W of power - enough for two incandescent bulbs. Top athletes can produce more - in short bursts. Your body is roughly 25% efficient in converting "calories" (which are actually kilo calories) to Joules - meaning that if you work out hard enough to burn 600 kcal per hour, then you actually produced ...

2

A (kilo)calorie is a unit of energy, while a watt is a unit of power, which describes the rate at which energy is expended. So a 100W bulb is using 100 joules a second. A kcal is about 4184 joules, so a 100W bulb takes about 42 seconds to consume (really: convert into light and heat) a kcal. The joule is the SI (derived) unit of energy. Units of energy ...

2

An average person uses approx. 1500-2500kcal/day. Since one kcal equals 4148J in SI units, that's between 6.2-10.4MJ per day. A day has 86400 seconds, which brings us to an average power consumption of 72-120W... about as much as a light bulb. :-) Physical exercise varies between light (300kcal/h) at an additional 350W to very strenuous at probably six ...

1

You're not the first, nor the last, to find the phrase "power flow" somehow wrong. For example, from W J Beaty's article on electrical misconceptions: ELECTRIC POWER FLOWS FROM GENERATOR TO CONSUMER? Wrong. Electric power cannot be made to flow. Power is defined as "flow of energy." Saying that power "flows" is silly. It's as silly as saying that ...

0

Well, if you think about electric power, which includes current (notion of flux), then you'll end with the conclusion that if there's no flow, there is no power.

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It's true that there are many inner products you can choose on $\mathbb{R}^3$. However, physics supplies the additional principle of rotational invariance: the result should not depend on our coordinate system. Now, any inner product of vectors $a$ and $b$ can be written as $$a \cdot b = a^T M b$$ for a matrix $M$. Rotational invariance tells us that $M$ ...

3

There is a little bit more thinking behind saying that $P=\vec F \cdot \vec v$ than it being a generalised multiplication in 3D. There are even cases where multiplication with scalar becomes a cross product when using 3D vectors. For example, torque $T=Fr$, becomes $\vec T = \vec r \times \vec F$. Whenever implementing vectors into existing scalar equations, ...

0

It’s all about engine's efficiency. According to wikipedia: gasoline engine's efficiency = 1/(BSFC × 0.0122225) (Also Actual efficiency can be lower or higher than the engine’s average due to varying operating conditions.) To calculate BSFC (Brake specific fuel consumption) use the formula: BSFC=r/P (where: r is the fuel consumption rate in grams per ...

3

First answer this: does it take more power to run one light bulb or two? Transmission lines are designed to be low loss, but they run a long way. Lightbulbs are designed to be "lossy" because that's how they work. Let's say for the sake of argument that the whole transmission circuit loses 100 watts and you're talking about a 100 watt bulb: those numbers ...

2

I believe the main effect of Earth Hour is in attracting public attention to sustainability. But besides of it, without doubt it saves energy (... and natural resources as well as one's wallet) if one turns off their unused light. The scale of one hour savings on turning off dispensable lighting for one hour is however rather small compared to other means ...

0

When you are asked the average of power, we do not care about how the power is fluctuating over time. All that concerns us is the amount of power delivered over time, and then we divide by that time duration. Whereas when it's instantaneous, we have to be more careful. We have to know the rate at which power is being delivered. ( This is often calculated by ...

2

How can you do that when Vs and Vd are two different parameters? One must keep track of the variables. The power delivered to a resistor is $$P_R = V_R \cdot I_R = V_R \left( \frac{V_R}{R} \right) = \frac{V^2_R}{R}$$ where I have subscripted the variables so it is clear that the voltage and current variables are the voltage across and current through ...

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P = Vs I and P = Vd I , both are correct. Former means Power consumed by circuit and latter power consumed by different circuit components like wires. Or $V_{s} I = V_{1} I + V_{2} I ...... V_{n} I$ Where V1 and Vn are voltage across different circuit components. If you use $P_{s} = V_{s} I$ and $V_{s} = IR$ You get , $P_{s} = \frac{V_{s}^2}{R}$ which ...

1

Yeah, I had this confusion too. But know that$$P=\frac{v^2}{R}$$ for RESISTOR circuits only. Actually Power for any circuit is (Instantaneous power more precisely) $$P=VI$$ Here's how: We know that $P=\frac{dW}{dt}$ Lets first calculate dW. dW is the elemental amount amount of work done on elemental charge dQ in moving it through a potential difference of V ...

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I think that by Kirchhoff's voltage law the voltage drop $V_d$ must equal the supply voltage $V_s$. So $V_d=IR=V_s$ and hence the result.

2

There are several interacting effects. And for global warming, we need to look at the whole earth. First effect: putting the panels up may change the albedo - the reflectivity - of the surface. That will increase the amount of energy absorbed by the earth, rather than reflected into space. The effect would be a miniscule increase in heating. Second ...

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The energy budget of the earth is predominantly dependent on the balance between radiation coming from the sun ( there exists internal energy from the magma in the earth but it is a small percentage of the energy budget) and radiated energy from the earth back to space according to a modified black body radiation. There is a secondary energy budget that ...

0

Solar panels are dark (they have to absorb light) and they do decrease the Earth's albedo, i.e. they make the planet locally darker. This means that locally they will cause slight additional heating. Part of the energy that a panel absorbs gets converted into electrical energy, which can be transferred off-site, i.e. removal of energy actually cools the ...

3

First of all the Earth's radius is about $6.4\times 10^6\mathrm{m}$, while its orbital radius is about $1.5\times 10^{11}\mathrm{m}$: about $2.3\times 10^4$ times its radius: yes, it is a good approximation that the whole Earth is at the same distance from the Sun. Secondly, this is the flux passing per unit area of the imaginary shell: the flux per unit ...

2

The only lossy component in the circuit is the resistor and the power dissipated in the resistor is $I_{\text{rms}} R$ so the conversion $I_{\text{rms}} = \dfrac {I_{\text{peak}} }{\sqrt{2}}$ has to be made assuming that the variations of current and voltage are sinusoidal. The mean power dissipated in the capacitor over a cycle is zero.

2

The difficulty is that there are three voltages involved. The voltage at the power station end $V_S$, the total voltage drop across the cables $V_L$ and the voltage at the consumer end $V_C$. The voltages are related as follows. $V_S=V_L+V_C$ So you have power supplied by power station is equal to the power lost in the transmission cables plus the power ...

0

Yes, your answer is correct, except You forgot to divide I by √2. $P=I^2 Z$ $P=I^2 |Z| cos \phi$ $P=I^2 |Z |\frac{R}{|Z|}$ $P=I^2 R$

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Power consumer and power dissipated will be different. Power consumed takes into account power supplied to the resistor and the capacitor. Power dissipation is the I^2.R losses due to heat. The power consumption of the capacitor falls on the imaginary axis. Hence we never observe it as heat. Here's a guy asking a similar question: ...

1

As you have confirmed the markings on the batteries to be, on one hand, 7.4V 2000mAh and on the other, 11.1V 1400mAh, and also considering that I have no knowledge about heated gloves, you can think of it as follows: The 7.4V 2000mAh battery can deliver, say, 200mA for 10 hours at approximately 7.4V. I say can because what determines the actual current, is ...

1

I read some of your comments. Explaining why $H = I^2R$ is not working out If same power is supplied for same time, equal heating would take place. $H = I^2R$ is not working here because R in two cases are different. $V = IR_{1}$ $7.4 = 2 R_{1}$ $R_{1} = 3.7 ~\Omega$ Similarly , $R_{2} = 11.1/1.4 = 7.928 ~\Omega$ Now, if you use $H = I^2Rt$ ...

1

Are you sure the batteries are labelled "2 A" and "1.4 A" rather than "2 Ah" and "1.4 Ah"? A battery doesn't have a fixed intensity. It has an approximately fixed voltage, and the intensity depends on the resistance of the connected load. So usually batteries bear two numbers: voltage and capacity (usually expressed in ampere-hour, 1 Ah=3600 coulombs). If ...

1

More the power, more the heat generated. $$P(\textrm{Power})=V\cdot I$$ $$H(\textrm{in joules})\propto V\cdot I\cdot t\,(\textrm{in s})$$ $$\therefore H(\textrm{in joules})\propto P\cdot t (\textrm{in s})$$ The 11.1 volt gloves does not give greater heat just because it has more Potential difference, but because $\textrm{Power}$ is greater in it.

2

What will happen, is that any three-phase AC motors powered locally will reverse their rotation direction. That might break some items, will definitely inconvenience folk with right-handed twist drills, when their drill press, under power, can only back out of the hole. Table saws will aim their gums at the cut instead of their teeth. Big fans will stop ...

1

Assuming a linear electrical circuit with two or more sources, we can always 'decompose' the current through a wire into 'component' currents due to each source; this is the superposition theorem. However, since the brightness of the bulb depends on the power delivered, and since power is quadratic in the voltage or current for a resistive element (such as ...

0

There isn't any common physical scenario where it makes sense to say "I have two currents flowing in opposite directions in a wire". In mesh analysis we may say the current through a particular wire is the sum of two circulating mesh currents. But we mean the current through that wire is just a single value, determined by summing two values. We don't mean ...

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