Is there a specific formula where I can compute how many hours before my 12v car battery becomes fully empty when using 100w light bulb only?
78096
How long can I run a water pump which requires 12V and 50W from a 12V, 40Ah, 400A car battery?  Can you explain how this was calculated?
37359
I need to create a task in TS based on battery level change. Let's say my battery drops from 67% to 66%. How can I run a task based on this event. Does Windows log this at all? I couldn't find this information anywhere.
16922
I am building a battery powered device. Almost all of my logic is running at 3.3v, although I do have a 5v supply for a few hobby servos.  I want to measure the battery voltage, but I can't do so directly with the ADC on my controller because the battery voltage (depending on the cell count of the lithium polymer battery that the user connects to the device) can vary between 6 and 36 volts. (My 5V and 3.3V supplies are fine with this input variation already.) The ADC can't measure anything above its supply voltage.  Obviously I need some sort of amplifier with a fractional gain, about 1/12 or so. Should I just use a simple voltage divider? How do I decide what values to use? Maximizing sleep-mode battery life is a concern, so that votes for using large values. What constrains how large those values can be? (The input impedance of the ADC, right?)  Am I better off using some sort of opamp amplifier circuit for some reason?  What's the usual solution here?
110320
I haven't found anything yet so I was hoping maybe someone here had an idea.  I want to kick off an even based on battery on an Apple laptop reaching a certain charge level, or when the battery is fully charged.  I know I can have something run periodically, check the battery level, and decide what to do, but I prefer event driven if possible.
1568
I understand that I can not connect an LED directly to a battery because it will draw too much current. Thus, there must be something else in the circuit to limit the current. What options are there? Are some methods more efficient than others?
39044
Two sources of power:   12V switched power supply (primary) 12V deep cycle lead acid battery (secondary)   Note: In practice the switched power supply voltage will be adjusted to match the actual voltage across the battery when it is connected to the load.  12V Relay:   NO connected to secondary power supply NC connected to primary power supply Relay coil connected to primary power supply   Load (12V):   Surveillance cameras Network attached storage Router Assume Total 10A   Simplifications:   safety precautions such as fuses ignored for now, battery Ah, run time etc. ignored for now, why am I doing this ignored for now.   Question:   When the relay switches from one power supply to the other there will be a temporary open circuit.  How do I calculate capacitor size to use? The open circuit time is unknown but can I assume it is say 10ms?   Solution? Charge:  $$Q = 10\text{ms} \cdot 10\text{A} = 0.1\text{C}$$  $$Q = C \cdot V$$  Capacitor size:  $$C = \frac{Q}{V} = \frac{0.1}{12} = 8,333 \mu \text{F}$$  Now I realize that this means that at the end of 10ms the Voltage is zero. So I figured I'd ask for some advice:)
41841
Consider two circuits containing a battery, bulb, switch and conducting wires but of length 1 meter in one case and 1000KM in other. When switched on at the same time both the bulbs glow instantaneously. I think it is because, electric filed travels from positive terminal of battery to negative terminal through wire almost instantaneously.  My question is there any speed for this electric field? If yes, what is it? By any modification in this circuit is it possible to induce any delay in glowing the bulb?
27887
How do I calculate the approximated time for the Charging and Discharging of the battery? Is there any equation available for the purpose? If yes, then please provide me.
78706
I'm new to electrical engineering, and probably asking the wrong question. Not sure how else to phrase it.  For the question, I'm going to use an example layout of simply a 9v battery attached to a toy DC motor.  From my understanding of my , the voltage coming from the battery through the motor will be around 9v. The current will be from 0amps to however much the battery can supply without frying. What decides how much current goes through the motor?
75428
What I would like to do is have an LED turn off 1 minute after the power is turned off. The LED would be on all the time when power is on. Thus:  Turn power on: LED comes on immediately. Turn power off: LED stays on for 1 minute and then turns off.  This is to signal when someones eyes have adequately adjusted to the dark in a darkroom.  Thanks in advance.
87864
Instead of looking up , how can I calculate Lux from EV?
71080
I chose an LED for my application but I need optical power, in mW.  In the , luminous intensity is shown in mcd but I did not find formulas for calculation although there are graphs, etc.  Where luminous intensity is 180 mcd and viewing angle 120°, etc.  Please provide any necessary formulae for my case.  I could not find the exact formula on this website.
55685
When t&lt;0 Components in circuit are Jt = 4 A, R1 = R2 = 2 Ω, R3 = 4 Ω ja L = 10 H      At t=0 switch K will be closed and i need to figure out what iL(t) is when t=4  So i tried to figure out iL(t) with  differential equation. First i combined R1 and R2 to get R12=1Ω  Then i transformed power supply from current to voltage Et=Jt*R12=J*1Ω =J Then i combined R12 and R3 to get Rz=5Ω Now i can create equation which is  $$E_t=L*\frac{di(t)}{dt}+R_ti(t)          $$ After adding constant i get $$J_t=10*\frac{di(t)}{dt}+5i(t)          $$ First i found the roots $$10r^t+5r=0 -&gt; r=-\frac{1}{2}$$ So for homogeneous part i get $$y^h=C_1*(-\frac{1}{2})^t$$ And for nonhomogeneous part $$J_t=A$$ $$ \frac{dJ_t}{dt}=0 , \frac{d^2J_t}{dt^2}=0$$ So i get $$  10*0+5=A-&gt;A=5  $$  For complete equation i get $$ y(t)=C_1*(-\frac{1}{2})^t+5  $$ To figure out what is C1 i use t=0 $$ y(0)=C_1*(-\frac{1}{2})^0+5=4 -&gt;C_1=-1  $$ And now for t=4 i get  $$ y(4)=-1*(-\frac{1}{2})^4+5=4.5   $$ I know this answer isn't correct but i don't know what went wrong Also sorry if its little bit hard to read this, i'm new to this
74403
If I had a 12V battery and wanted to use an LED that used 2V and 20mA, then I need a resistor that will drop 10V and provide enough resistance to produce a current of 20mA. Therefore I would use R=V/I = R = 10/0.020 = 500 Ohms.  Therefore I would need a 500 Ohm resistor. I believe this is correct so far?  My confusion comes from when we then look at the circuit as a whole. The circuit has a current of 0.020A and a resistance of 500 Ohms, which when looking at V=IR equals 10V (the resistor), not the 12V that it should be there for the whole circuit?  Where am I going wrong? How do I make V = IR work regarding the whole circuit so that V = 12?  I'm pretty new to this sort of thing and no doubt I've probably made a daft mistake but if someone could just help clear up my confusion it would be appreciated.  Thanks.
115199
I'm using a 850nm IR LED (), and in order to extend my battery life, I'd like to make the LED blink (UF 1.9 V, IF 100 mA).   I found a couple of links , ,  but they either a) require two LEDs that blink in alternation, or a bunch of resistors that will most probably just drain my battery even quicker (will they?).  Therefore my question:  What's the    most simple energy efficient  smallest possible (ideally not bigger than the battery &amp; LED) ideally operated with 3V or less.   circuit to make an IR LED blink?  EDIT:   I'd like to blink the LED 10 to 30 times per second. But We can start with 2 to 5 times. the LED should be on for 10ms (at least) or 100ms (at most). Depending on blinking frequency. I'm trying to make the circuit as small as possible (size and complexity) cause I wanna use it for a tracking device. I'm powering the LED with a 3V CR2032 battery and I was hoping I could extend the battery life by making the LED blink instead of having it on all the time. I can currently run the IR LED with the CR2032 battery for 4 hours. Extending that time to 8 or maybe even 12 hours would be great.
54246
I would like to get an alert (with sound) when my laptop battery is fully charged.  How can I do that?
45577
I've implemented above circuit and it was successful.  The given circuit needs 12VDC to run properly.  I want to run this circuit on 5-6VDC or 9VDC whichever is possible. I know I have to change resistors but not confirmed which to use.  Also if I use 9V battery only to activate 6V relay for 8hrs daily, for how many days will that battery work?