Henryv111111
Member
Question If you connect a external resistor 12V to 6V should one side of the resistor to grd read 12V and the other side read 6V?
Thank you
Thank you
- Thread starter Henryv111111
- Start date
Indiana Ken
Well-known Member
The voltage drop across a resistor is a function of current (amperes) and is given by: Voltage Drop = I x R where I is in amperes and R is the resistance value of the resistor in ohms. For example, if the resistance value is 2 ohms and the current is 3 amperes the voltage drop acros the resistor will be 6 volts (2 x 3 = 6). In this case for a 12 V battery, the battery side of the resistor to ground will measure 12 volts and the load side of the resistor to ground will measure 6 volts.
This is the down side of using resistors to change voltage levels - it only works when the current is constant and does not change.
Henry, the answer is probably YES PROVIDED the ballast is the correct resistance to match the 6 volt coil and the points are closed. If the ballast is approximately the same resistance as the coil, you would have a 50 50 voltage divider so theres 6 volts dropped across the ballast leaving 6 for the 6 volt coil when the points are closed and conducting current. Many coils labeled 6 volts have around 1.25 to under 2 ohms primary resistance while a typical 6/12 ignition ballast resistor is in the same ohms range SO THE 50 50 6/12 VOLTAGE DIVISION IS ACCOMPLISHED.
Of course, the exact answer depends on the coil and ballast resistance and a good set of closed points but you are on the right track assuming its a typical 6 volt coil and typical 6/12 ballast system.
John T
Of course, the exact answer depends on the coil and ballast resistance and a good set of closed points but you are on the right track assuming its a typical 6 volt coil and typical 6/12 ballast system.
John T
old
Well-known Member
- Location
- Camdenton Missouri
Yes and no. It depends on if the circuit is open or closed as in in an ignition system if the points are open you will see 12 volts on both sides. If the points are closed you will see 12 volts on one side and 6 on the other side give or take a little bit
John M
Well-known Member
- Location
- Nunyafn business
I assume youre refering to from the switch to the resistor, not the coil, which is why I dont really understand but one answer, youll have 12 tothe one side of the resistor, 6 give or take on the other, shouldnt matter whether the points are closed or not at the resistor. would be 12 volt at the hot side of the coil not matter what if the resistor wasnt there.
John M
Well-known Member
- Location
- Nunyafn business
Bob, Im not taking about the coil, not do I think the OP is either. He said 12 volt at the resistor on one side, 6 volt on the other. (I can honesty say, Ive never checked the voltage on the "out" side of a resistor, other than to see if it was there, IE my meter moved, so 6 or so volts on that side may be off) BUT what Im saying is you have 12 volts at the battery, to the switch to the coil NO MATTER if the points are open or closed. Putting a resister in line doesnt change that 12 volts, take the wire off, you still have 12 volts to it, so why does it matter it the points are open or closed, as is mentioned by others.
because we live in a world governed by physical laws.
those physical laws state that current flows in a closed circuit, not an open circuit.
ohms law describes it the best .. e=ir
if you do not have current flow, you do not have forward voltage drop across a component like a resistor.
for instance, battery to swtich to resistor to coil, to ground.thru points.
if points are closed, then current is flowing inthe circuit, and the coil will show, if read by a meter, some voltage less than 12v on it, depending on the resistor value, and coil value.
ie.. 12v incoming from battery to resistor, and assuming equal resistance coil and resistor, then the resistor will drop 6v, and thus the coil only sees 6v.
now.. if points are open... everywhere you measure in that circuit is going to show 12v.. you could have a hundred.. or 1000 ressitors in line, and all will show 12v on them as long as current is not flowing.. ie.. points open... no current flow.. no voltage drop.
and yes.. i do have my engineering degree...
soundguy
those physical laws state that current flows in a closed circuit, not an open circuit.
ohms law describes it the best .. e=ir
if you do not have current flow, you do not have forward voltage drop across a component like a resistor.
for instance, battery to swtich to resistor to coil, to ground.thru points.
if points are closed, then current is flowing inthe circuit, and the coil will show, if read by a meter, some voltage less than 12v on it, depending on the resistor value, and coil value.
ie.. 12v incoming from battery to resistor, and assuming equal resistance coil and resistor, then the resistor will drop 6v, and thus the coil only sees 6v.
now.. if points are open... everywhere you measure in that circuit is going to show 12v.. you could have a hundred.. or 1000 ressitors in line, and all will show 12v on them as long as current is not flowing.. ie.. points open... no current flow.. no voltage drop.
and yes.. i do have my engineering degree...
soundguy
John M
Well-known Member
- Location
- Nunyafn business
But he said to ground so he should have 12 volts to the input side of the resistor right. The way I"m reading replies is that the points have to be closed to have voltage there. I don"t know one the output side of voltage as I already mentioned but on tje input side of the resistor there sjouls be 12 volts whether the points are open or closed if he"s going to ground right or has that changed.
yes and no.
you always show bat voltage to all connections in a open series circuit that are on the side tied to battery.
once the points are closed it is no longer a open circuit.
the reason you may not always have 12v to the side of a resistor in a closed series circuit is if there are other resistors ahead of it. IE.. a machine that uses a ballast resistor and has an added resistor for a 12v conversion. if points are closed, only the first resistor in line will have 12v on it.. the next resistor, will have less, and the coil will have even less. each device will drop voltage, in a closed circuit.
that's why you see peopl ask or specify whether points ar eopen or closed.
soundguy
you always show bat voltage to all connections in a open series circuit that are on the side tied to battery.
once the points are closed it is no longer a open circuit.
the reason you may not always have 12v to the side of a resistor in a closed series circuit is if there are other resistors ahead of it. IE.. a machine that uses a ballast resistor and has an added resistor for a 12v conversion. if points are closed, only the first resistor in line will have 12v on it.. the next resistor, will have less, and the coil will have even less. each device will drop voltage, in a closed circuit.
that's why you see peopl ask or specify whether points ar eopen or closed.
soundguy
John M
Well-known Member
- Location
- Nunyafn business
I understand that but the original poster said to ground on the voltage. So if he's going to ground woth his meter wouldn't je have voltage on both sides of tje resistor to ground. I'm sorry I'm just seeing why the points would have to be closed to get voltage to the resister and out of tje resister if he's going to ground. Also sorry for all my typing mistakes. Trying to do this on my phone at the doctors office.
you are missing the point..
ground is a reference location
one side of the meter is to ground, as a reference,..the other side .
lets assume a simple ignition ckt. 12v baqt, ign switch, 1 resistor, and the coil, pionts,, etc.
with points open, you get bat volts on both sides of the resistor and both sides of the coil..
with points closed you get bat volts on the sweitch side of the resistor, and some less volts on the other side of the resistor / input side to the coil, and ov course no volts on the points side of the coil
soundguy
ground is a reference location
one side of the meter is to ground, as a reference,..the other side .
lets assume a simple ignition ckt. 12v baqt, ign switch, 1 resistor, and the coil, pionts,, etc.
with points open, you get bat volts on both sides of the resistor and both sides of the coil..
with points closed you get bat volts on the sweitch side of the resistor, and some less volts on the other side of the resistor / input side to the coil, and ov course no volts on the points side of the coil
soundguy
John M
Well-known Member
- Location
- Nunyafn business
That's exactly what I'm saying.the poster asked about voltage at the resistor. So why would the points come into play for voltage at the resistor like other were saying. I guess I wasn't communicating that correctly.
at the resistor.. therein is the issue.
a resistor has 2 sides.. input and output.
in a 1 resistor setup, the input side is essentially the battery, and will always be at bat volts.. output side will change with forward voltage drop depending on if points are open or closed.
soundguy
a resistor has 2 sides.. input and output.
in a 1 resistor setup, the input side is essentially the battery, and will always be at bat volts.. output side will change with forward voltage drop depending on if points are open or closed.
soundguy
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