# How to transfer energy from a generator to a storeage battery

and thank you in advance for taking the time to read my question. To give an idea of my working level, I'm a 21 year old computer science student entering my senior year at college. It's been a few years since my Electricity/Magnetism course, and i'm a bit rusty on the Lorentz Force. I wanted to create a sort of "Human Powered Generator", ie. something as simple as a stationary bicycle turning a generator as I peddle. Now I know the "right hand rule", and can quite easily make a motor/generator with some wire and magnets. My question is about voltage/current. I was never clear on the effects of the strength of the magnet and its importance in the amount of potential power generated. In other words,

1.) If i'm turning a generator at a constant rate and it is lightning a bulb, if I magically replaced the magnets with ones twice as strong, what would happen? I'm assuming it gets twice as hard to turn, but outputs potentially twice as much power.

In addition, I was never clear on the relationship between voltage and current in the lorentz force.

2.) While turning a generator at constant speed S with magnets of strength B and # of wire coils C, how much voltage/current is created? I know there are many variables involved here, perhaps such as the width of the rod the coils are wrapped around, thickness of coils, etc.

Finally,

3.) If trying to charge a battery of V volts and A ampre-hours, what measures should I take to ensure safe delivery of energy to the battery? In other words, if I peddle the generator very rapidly, I expect lots of either current/voltage/both. I assume I need a voltage regulator of sorts, and I'm not sure if the current matters (I think its just how "much" energy there is, whereas voltage is the "pressure" or "strength" of the energy).

I appologize if any assumptions I made are incorrect, i'm just going off of old knowledge. I tried wikipedia, but its all symbolic and I cant find a hard example (As in, I dont know how to find magnetic strength of a magnet, always denoted as B in the equation). Thanks to anyone who can answer any/all 3 of my questions!

-John

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You would be unwise to try recharging batteries with a homemade system unless you know what you're doing. In the worst case, you can start a fire. More commonly, you'll reduce the battery's capacity and lifetime by damaging the electrodes. To do it properly you would need to build a little regulator circuit that is specifically geared to the battery chemistry. ...Ask an electrical engineer for help. –  Steve B Jun 4 '12 at 20:54
Indeed, I would not want to start any fires. Im preparing myself to do a small project and Pygmalion has given me a lot of knowledge on the subject. I will now go to learn about regulating voltages. Thanks for the warning though, and I will certainly ask an EE in person at college. Unfortunately, its summer break for us at the moment. –  John Toniolo Jun 5 '12 at 17:42

First a few lines of basics. If you put a loop into the magnetic field and this loop turns within it, the magnetic flux through loop shall change according to the formula

$$\Phi_B = \vec{B} \cdot \vec{A} = B A \cos\phi = B A \cos\omega t,$$

where $\vec{B}$ is magnetic field strength, $\vec{A}$ is area of the loop and $\phi$ is angle between $\vec{A}$ (perpendicular to loop) and $\vec{B}$, while $\omega$ is angular velocity of the rotation of the loop.

If you use coil with $N$ loops, then induced voltage on the coil shall be

$$\mathcal{E} = N \frac{\text{d}\Phi_B}{\text{d}t} = - N B A \omega \sin\omega t.$$

Therefore, yes, if you use twice larger magnetic field, you get twice larger voltage.

However, twice larger magnetic field does not necessarily mean it is twice as hard to turn. For example, if you have no electric load on the generator, there is practically no current in the coil and there is practically no Lorentz force! However, if you have completely ohmic load, voltage twice larger means current twice larger and yes it becomes twice as hard to turn.

In practice, you should also consider the internal friction of the generator. So even if there is no load, some muscular power will be required in order to overcome friction. When you increase the electrical load, required power in order to turn generator increases.

Of course, it is difficult to keep rotation constant, which means that with larger angular velocity $\omega$ voltage increases. To keep generator's output voltage constant, you need some electronic circuit, of which the simplest possible includes zener diode.

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Awesome answer, thank you very much! That clears up the first 2 questions. Do you have anything to say about 3? According to your equation, an increase in angular velocity (im meters/sec), magnetic strength, and # coils is directly proportional to the induced voltage. Im guessing I can find B by using a constant rotation and reading the voltage with a multimeter, unless there is a way to find the "strength of a magnet". Now with this information, I would just simply need a voltage regulator (something to transform a variable x volts to necessary constant Y volts for battery storeage) correct? –  John Toniolo Jun 4 '12 at 20:52
First of all, angular velocity is in radians/sec, $\omega = 2 \pi f$, where $f$ is frequency of rotation. All above equations are of course approximate, under assumption that permanent magnet is much larger than coil, so the coil is positioned in the homogenous magnetic field. In practice $B$ is not homogenous, but induced voltage should be still very well proportional to $N$ and $f$. So you change $N$ and $f$ accordingly in order to obtain approximately the correct voltage. You also need some kind of voltage regulator, especially since you cannot ensure that $f$ is constant in time. –  Pygmalion Jun 4 '12 at 20:59
Awesome thank you, you've been a great help! One last thing since you're so knowledgeable. You mentioned I should change N and f accordingly to obtain the correct voltage. Well, when I peddle harder, I want more "energy", even if limitnig the voltage. Is it still along the same order of efficiency to make less coils? In other words, given a constant angular velocity, and all else being constant except the number of coils, will the energy produced be the same? I beleive I remember that less coils means more current at a lower voltage (ie. transformer). Thank you so much again –  John Toniolo Jun 5 '12 at 13:13
@JohnToniolo The "problem" with power is that in general you cannot regulate power and voltage independently - larger voltages larger power. This is pretty obvious for pure resistive load, i.e. $P = U I = U^2 / R$. So you have to keep voltage constant and the load will take as much power as it needs. More power it needs, larger currents and harder to turn generator. It all makes sense in terms of conservation of energy. –  Pygmalion Jun 5 '12 at 16:26
Very informative, learning a lot. So if I was charging a dead battery (with a voltage regulator holding generator voltage constant), for instance, it would be hard to turn at first. As the battery became close to full charge, however, the generator would become easier to turn? And the quicker I peddle the generator, the quicker the battery charges? Or would peddling too fast blow it up? My old physics teacher demonstrated what you're saying with a small hand generator. When a bulb was not attached to the wires, it was very easy to turn. When it was connected, it was very hard to turn! –  John Toniolo Jun 5 '12 at 17:37