Hello few days back i serviced the starter motor of my motorcycle. brushes were partially worn but they were in spec according to the manual. i got this question while i was doing it. my starting system has been working fine. i checked the starting system with a multimeter after servicing my starter motor. i just needed to make sure everything was ok. otherwise there was no need to do it.

anyway when i test the starting system with the multimeter i can see the normal voltage drop which we see in both cars and motorcycles. every time i press the self start button the battery voltage drops from 12.8V to 10V (roughly). when i honk the horn i can see a slight voltage drop too. on my horn they have printed it as 12V 5A. so its amperage value is there. but on my starter motor (brand is MITSUBA) i can't see any amperage value. both positive and negative wires of the starter motor is very thick. so apparently it draws more amps.

but my questions are,

1) what makes this voltage drop. does more amperage draw always mean a voltage drop ? since voltage is the potential difference how potential drops as the current (amps) rises through the circuit ?

2) what makes the starter motor to draw more amps than the motorcycle horn? is that the resistance inside the device (motor) which directly affects the amperage? if we apply V=IR we can see that as we increase the resistance the amp value drops. if we decrease the resistance the amp value rises. but that's assuming the voltage is constant. so i'm not sure it's practical to apply it for this example ( sorry i have a very basic knowledge ).

3) is there a way to check the amp value of the starter motor? i have used volts,ohms,diode and continuity modes in my multimeter but haven't used it to measure amps so far.

thank you very much

 

The internal resistance of your battery.
Vout = Vin - IR
10V = 12.8 - IR
2.8V = IR

You need to find R to figure out I. Since you do not know I, the amperage rating of the starter, you need to find out R, the internal resistance of your battery.

You can find R roughly by measuring the exact voltage drop across your 5A horn and use it to find your starter rating.

 

A starter motor has a very low impedance as it requires a lot of power (in the neighborhood of a horsepower) to start a motorcycle engine.
Thus a motorcycle starter (depending upon he size of the engine) draws in the neighborhood of 50A which would correspond to an impedance of about 0.2 ohms.
That is likely higher than the measured winding resistance as the motor generates some back emf when running which reduces the current.
That value is too low to accurately measure with a standard multimeter.
To measure such a value you could pass a known current (such as 1 ampere) through the motor and measure the voltage drop.

This high starting current causes the voltage drop you observed in the battery internal resistance, which is apparently only about 0.05 ohms.

 

...my questions are,

1) what makes this voltage drop. does more amperage draw always mean a voltage drop ? since voltage is the potential difference how potential drops as the current (amps) rises through the circuit ?

Resistance
Voltage = Current x Resistance
The more current you draw, the more the voltage drops.

In an ideal world, there is zero resistance in the battery and zero resistance in the cables. The battery voltage would stay the same regardless of how much current the load takes. In the real world, the battery has internal resistance. On top of that, the cables have resistance too. It depends on where you measure the voltage. If you measure the voltage at the battery then you will witness the internal resistance of the battery. If you measure the voltage at the load then you have to include the resistance of the cabling.

2) what makes the starter motor to draw more amps than the motorcycle horn? is that the resistance inside the device (motor) which directly affects the amperage? if we apply V=IR we can see that as we increase the resistance the amp value drops. if we decrease the resistance the amp value rises. but that's assuming the voltage is constant. so i'm not sure it's practical to apply it for this example ( sorry i have a very basic knowledge ).

The difference is in the power required by the load, i.e. the motor or the horn. The horn does not require as much power as the motor. The winding in the horn will have much higher DC resistance.
The winding in the starter motor will use much thicker wire and will have lower DC resistance. Moreover, a motor is more than just a resistor. It is also an inductive load. When the motor starts up it requires a very large current initially. As the motor begins to turn the current will fall.

3) is there a way to check the amp value of the starter motor? i have used volts,ohms,diode and continuity modes in my multimeter but haven't used it to measure amps so far.

The standard way of measuring load current is to break the connection and place an ammeter in series with the load. However, most DMM can only measure up to 20A. To measure a higher current, one can insert a very low resistor in series, for example a 10 milli-ohm resistor capable of carrying 100A, and measure the voltage across the resistor. In other words, you need an ammeter capable of measuring 100A or more.

 

Things that store energy can be characterized as a perfect <whatever it is> combined with an equivalent <whatever its imperfections are>. A battery is considered a constant-voltage source. That means that a perfect battery has the same terminal voltage no matter what the output current is; 1 mA, 1 kA, don't care.

In the real world, nothing is perfect and anything that conducts electric current has an impedance of some kind (we'll leave out superconductors for now). So a real world battery is a perfect battery in series with an impedance, and for a battery the vast majority of that impedance is resistance (as opposed to inductance), called its ESR (equivalent series resistance). Most electrolytic capacitors also have ESR specs. In a high-capacity battery such as in cars and bikes, the ESR is very small but very real. If you consider a battery as a battery with a resistor in series, they you can use Ohm's Law to experimentally determine the ESR. Note that ESR is not a nice, well-behaved constant. Its value changes somewhat with the amount of current being drawn from the battery, as well as with temperature.

ak

 

The internal resistance of your battery.
Vout = Vin - IR
10V = 12.8 - IR
2.8V = IR
You need to find R to figure out I. Since you do not know I, the amperage rating of the starter, you need to find out R, the internal resistance of your battery.
You can find R roughly by measuring the exact voltage drop across your 5A horn and use it to find your starter rating.

got it thx. yea as you said we can use our horn to calculate the R.
but are battery internal resistance and the circuit resistance two different things ? i mean are the resistance of the relevant component (motor or horn) and the battery internal resistance two different resistances ?
is the R in this ( Vout = Vin - IR ) formula referring to the battery internal resistance ? for instance if we get the V=IR formula, is the R here also referring to battery internal resistance or the total resistance of the circuit ?
thank you

A starter motor has a very low impedance as it requires a lot of power (in the neighborhood of a horsepower) to start a motorcycle engine.
Thus a motorcycle starter (depending upon he size of the engine) draws in the neighborhood of 50A which would correspond to an impedance of about 0.2 ohms.
That is likely higher than the measured winding resistance as the motor generates some back emf when running which reduces the current.
That value is too low to accurately measure with a standard multimeter.
To measure such a value you could pass a known current (such as 1 ampere) through the motor and measure the voltage drop.

This high starting current causes the voltage drop you observed in the battery internal resistance, which is apparently only about 0.05 ohms.

i have a 30A fuse in my starter relay which also acts as the main fuse of the harness. so i think the motor draws a current which is lower than 30 amps.
in this case does low impedance mean low resistance ? or are they two different things ?
yea since the motor generates a back emf the resistance can vary. so as you said if we apply a known current and measure the resistance by looking at the voltage drop, are we going to use this Vout = Vin-IR formula ? in this case it measures the resistance of the motor. but as in @iimagine 's example it measures the battery internal resistance. so how do we decide whether it's the battery internal resistance or the total resistance of the circuit. sorry may be i understood it incorrectly. thank you

Resistance
Voltage = Current x Resistance
The more current you draw, the more the voltage drops.

In an ideal world, there is zero resistance in the battery and zero resistance in the cables. The battery voltage would stay the same regardless of how much current the load takes. In the real world, the battery has internal resistance. On top of that, the cables have resistance too. It depends on where you measure the voltage. If you measure the voltage at the battery then you will witness the internal resistance of the battery. If you measure the voltage at the load then you have to include the resistance of the cabling.



The difference is in the power required by the load, i.e. the motor or the horn. The horn does not require as much power as the motor. The winding in the horn will have much higher DC resistance.
The winding in the starter motor will use much thicker wire and will have lower DC resistance. Moreover, a motor is more than just a resistor. It is also an inductive load. When the motor starts up it requires a very large current initially. As the motor begins to turn the current will fall.


The standard way of measuring load current is to break the connection and place an ammeter in series with the load. However, most DMM can only measure up to 20A. To measure a higher current, one can insert a very low resistor in series, for example a 10 milli-ohm resistor capable of carrying 100A, and measure the voltage across the resistor. In other words, you need an ammeter capable of measuring 100A or more.

thank you got it.
would like to try the last method. when we place a 10 mili-ohm resistor in series and measure the voltage across the resistor,

it's going to give the voltage just across the resistor. but don't we need the resistance of both motor and the resistor (total resistance) to calculate the current ? i assume we are going to use V=IR here. thank you

Things that store energy can be characterized as a perfect <whatever it is> combined with an equivalent <whatever its imperfections are>. A battery is considered a constant-voltage source. That means that a perfect battery has the same terminal voltage no matter what the output current is; 1 mA, 1 kA, don't care.

In the real world, nothing is perfect and anything that conducts electric current has an impedance of some kind (we'll leave out superconductors for now). So a real world battery is a perfect battery in series with an impedance, and for a battery the vast majority of that impedance is resistance (as opposed to inductance), called its ESR (equivalent series resistance). Most electrolytic capacitors also have ESR specs. In a high-capacity battery such as in cars and bikes, the ESR is very small but very real. If you consider a battery as a battery with a resistor in series, they you can use Ohm's Law to experimentally determine the ESR. Note that ESR is not a nice, well-behaved constant. Its value changes somewhat with the amount of current being drawn from the battery, as well as with temperature. ak

thx a lot. never knew this. learned something new

 

R1 = Battery internal resistance -> Assuming 0.02
R2 = 12.8V / 5A = 2.56
Vout = 12.8V * (R2 / (R1+R2))
Vout = 12.8V * (2.56 / 2.58) = 12.7V
Thus, we can find R1 if Vout is known by measurement:
12.7V = 12.8V * (2.56 / (R1 + 2.56))

 

View attachment 185042
R1 = Battery internal resistance -> Assuming 0.02
R2 = 12.8V / 5A = 2.56
Vout = 12.8V * (R2 / (R1+R2))
Vout = 12.8V * (2.56 / 2.58) = 12.7V
Thus, we can find R1 if Vout is known by measurement:
12.7V = 12.8V * (2.56 / (R1 + 2.56))

thank you very much. now i got it

 

by the way hope it's ok to ask this here
when we calculate these kind of basic things we assume that the current (I) is constant through out the circuit.

but does this really happen in a real DC circuit. since current is the rate which electrons flow past a point, isn't this resistor (R) going to slow down that rate ( coulomb per second ) ? thank you

 

There are two components to "rate", quantity and speed. A resistor changes the quantity, not the speed. This is where the water analogy falls apart, that whole "speed of light is a constant" thing. At the same voltage, fewer electrons get through a large resistor compared to a small resistor, in the same amount of time.

Note that the velocity of the electromagnetic energy wave and the speed at which individual electrons are moving are two very different things. Back to the water analogy, a wire is like a hose that is full of water. When you open the spigot, water comes out of the far end of the hose almost immediately. BUT, it takes many seconds for that first drop of water going through the open spigot to get to the far end of the hose.

ak

 

There are two components to "rate", quantity and speed. A resistor changes the quantity, not the speed. This is where the water analogy falls apart, that whole "speed of light is a constant" thing. At the same voltage, fewer electrons get through a large resistor compared to a small resistor, in the same amount of time.

Note that the velocity of the electromagnetic energy wave and the speed at which individual electrons are moving are two very different things. Back to the water analogy, a wire is like a hose that is full of water. When you open the spigot, water comes out of the far end of the hose almost immediately. BUT, it takes many seconds for that first drop of water going through the open spigot to get to the far end of the hose.

ak

thank you for the explanation