Dreaded #2 Fuse blows immediatly in TC40D

[ccsial]

A starter solenoid draws by far the most current with the plunger all the way out (in the initial resting position). When current is applied from the starting circuit and the plunger moves in, the inductance goes up the current starts to fall.

FYI.

Since these solenoids work on DC. I believe you'll find the inductance has no affect on DC current. The coil resistance is the only factor on a DC solenoid.

Inductive reactance (XL) is the resistance seen by an AC current.
XL=2xPIxFreqxInductance. If Freq is zero (DC) then XL=zero leaving only coil resistance to impede current flow.

 

[SteveInMD]

FYI.

Since these solenoids work on DC. I believe you'll find the inductance has no affect on DC current. The coil resistance is the only factor on a DC solenoid.

Inductive reactance (XL) is the resistance seen by an AC current.
XL=2xPIxFreqxInductance. If Freq is zero (DC) then XL=zero leaving only coil resistance to impede current flow.

Good point. So it must simply be the fact that the pull-in coil draws far more current than the hold-in coil which causes the fuse to blow. If the pull-in coil is active too long - pow! I'm glad we have a solid explanation of what's happening here. These solenoids require full voltage to operate with reliability. However as long as you give them that they work just fine.

 

[barbqranch]

FYI.

Since these solenoids work on DC. I believe you'll find the inductance has no affect on DC current. The coil resistance is the only factor on a DC solenoid.

Inductive reactance (XL) is the resistance seen by an AC current.
XL=2xPIxFreqxInductance. If Freq is zero (DC) then XL=zero leaving only coil resistance to impede current flow.

Almost true. When you first close the switch, it is not unchanging voltage, but the induction opposes the inrush current, meaning it takes (a very brief) time to build up to full current.

Yes, I agree that parallel windings in play until the solenoid is fully engaged is the most likely problem.

 

[ccsial]

Almost true. When you first close the switch, it is not unchanging voltage, but the induction opposes the inrush current, meaning it takes (a very brief) time to build up to full current.

Yes, I agree that parallel windings in play until the solenoid is fully engaged is the most likely problem.

You are correct. I ignored the time constant (Tc=L/R) for this discussion.

 

[ccsial]

Jinman's explanation that there are two sets of windings in the starter solenoid might explain why a low battery could cause it to blow the fuse. If the voltage is low and it doesn't pull in far enough to disengage the second starter solenoid winding, then it would pull more current than it would with a full battery.

I suspect that is exactly right.