Can I connect 2 solar panels in parallel if they have different Open Circuit voltage ( e.g 22V and 14V)?
545877
Likely you can not connect them in parallel and may risk damage if you do.  Consider the analogy between electric current and water flow.  Voltage is equivalent to water pressure and electric current is equivalent to water flow rate.  If you connect two tall water towers in parallel, water will flow from the taller tower (higher voltage) into the lower tower (lower voltage).  You will likely drive electric current the wrong way through the 14V unit perhaps causing resistance heating and damage.
1082346
The "60 each" part of your question says you don't understand the most basic aspects of electricity.  Voltage is a measurement between two points.  Think of a 12V car battery.  It is 12 volts between the 2 terminals of the battery.  Another analogy:  two things that are 3 feet apart are 3 feet apart, not 1.5 feet each.\n\nDoubling 120 is 240, not 220, but I 220 is a commonly used number, like 110 for 120, but the standard voltages are 120 and 240.\n\nI think the point about doubling when you lose a neutral is talking about what is called a multiwire circuit.  The service into your house is the same, and some appliances like your dryer, have the same thing.  For a single phase case, a multiwire circuit has 2 phase (or hot) conductors and a neutral, with 120 volts between each phase and the neutral and 240 volts between the two phases.  This is created by a secondary winding on the utility transformer that is 240V, with a tap in the center which is grounded.  The tap is half way so it is 120V to each hot conductor.  If the neutral connection is lost, you still have 240V between the hot conductors, and all the 120V things in your house will be put in series.  For example, a couple light fixtures on 2 separate circuit breakers might be on the two opposite phases.  When the neutral is lost to the utility, there is still a connection of the neutral's of the lights, so you'll have the two lights in series and connected to 240V.  Each light will not have 240V though.  If the load on each phase were the same, removing the neutral will still leave 120V on each side.  That isn't likely, so one phase will get more than 120V (and generally cause damage) and the other phase will be less than 120V, and they will add to 240V.  It would be impossible for it to be so unbalanced that one phase would be 240V and the other would be 0.
606166
depends on the voltage
686453
u can use switchs to power each one at a time but u ll need to place them in a parallel format cause using serial means u have to power the first bulb to be able to light any other ,other than that u can use relays and connect them in a way to have the light going in a sequence
192726
No, that's not a good idea - neutral may not be the same as ground (0 volts).\n\nThink of it this way - the neutral wire provides a return path for the electricity to get back to the power company.  The ground is there to act as a shield, to protect the device from radio interference (if it's an electronic device like a computer).  If there's a ground fault, and you connect neutral to ground, the device may still work OK, but there may start to be "flakey" intermittent failures, like bad signal or data loss, because the neutral voltage is sending interference signals throught the device ground, instead of being shielded by the ground.
1038968
What voltage dc, or do you need to be able to vary it?  What current capacity do you need?  What ripple voltage is acceptable?
923631
Yes.You have to appropriately reduce the rating and use it at low voltage low current applications.
253109
I'm not sure but here you have some documentation!!!\nHigh Power Synchronous Buck Converter Delivers Up to 50A\nhttp://www.linear.com/pc/downloadDocument.do;jsessionid=Cj11DMriV8LW4X9CgZfQ5I5lx1vZBXzB2tjeDRsjX12XSI6CBFqH!-1570396240?navId=H0,C1,C1003,C1042,C1032,C1092,P1551,D4362
536506
Perhaps a clearer question could help others to answer your question correctly. Well, all I can do is supply you with this link: http://electrosofts.com/parallel
368541
Probably yes.\n\nFor general purposes, any voltage from 110V-130V can be considered the same thing.  Same for any voltage from 208V-240V.  The variations are due to many factors including distance from the outlet to the nearest transformer, capacity of the circuit, other attached devices, etc.\n\nMost devices today are capable of handling the seemingly wide variance in voltages. (e.g. a device rated for 220V can handle any voltage from 208V-240V.)
308968
if you are referring to creating electricity from uv light then any standard photovoltaic panel will do the job.  Depending on what photovoltaic material you use the efficiency will change, but because much UV is fairly close to the visible spectrum most will give you decent collection.
424895
It seperates wear the wattage is going between the two seperate rails... It is like a house wired for 240.  Some of your outlets go to one rail in the electrical panel on 120 some to the other rail of 120.  Then yet others use both rails for 240.  In the computer power supply they never really combine both rails though, just split where the end of the line goes.
644810
Your circuit has 6 resitance R1, R2, R3, R4, R5, R6 and correspond currents are: I1, I2, I3, I4, I5, I6 and voltage are: U1, U2, U3, U4, U5, U6.\nYou can make many kind of circuit(direct, parallel or both) depend on total voltage as well as resistance.
688170
Not sure. I use to be an electronics Tech but that was before SMPS existed. I'm guessing you might find something on the link below:
525922
Before you can answer this you need at lease to variables. You have the voltage of 1.5 vdc. You will need Amp the clock pulls. Wattage = Volts X Amps.
679488
simple use two resistors, 6 and 3k ohm , in series and take the volatage across the 3 ohm resistor, it will be 3V.....\n\nuse the laws,\n\nif you have a voltage source, and two resistors p and q in series with it, \n\nthe voltage across p is given by  Vx(p/(p+q)), here u know voltage across p(Vp)= 3V, or 4.7 V or whatever....\n\nso, Vp = V x p/(p+q)....\n\nu know the value Vp, and V so substitute them......\n\n3 = 9 x p/(p+q)\n\nso p + q = 3p, \n\nso "2p = q", take two resistors ( in kilo ohms, coz u cant use just ohms, coz they are very large) such that one is double the value of other.......\n\nand connect the circuit across the 'p' resistor......\n\nwith 4.7, u get,\n\n4.7 = 9 x p/(p+q).....solve ...and choose resistors accordingly
1112995
You need to hook enough solar panels in series to give a voltage slightly higher than that delivered by the car batteries.   I'll assume that they're 12 volt batteries.   Connect the panels in series until the output is about 12.5 volts.   Connect the ends to the terminals of your battery, checking that the polarity is correct.   When the batteries are partly charged, they'll easily deliver 120 VA for short periods, but unless you've got a very large array of solar panels, you won't get anything like this amount of power from the panels alone when the batteries are flat.   Car batteries are designed to deliver well over 1000VA to the starter motor.   I live in the S of NZ.   Not far from here there's a hut on a sheep station with a system like this that delivers power to a 12 volt lighting system.We probably get much less sunshine than you do in your climate, but last I heard, the system works fine.