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Ironically, it is also the most useful test procedure for)Tj T* ( locating faults and potential faults.)Tj 0 -2.375 TD (Simply, voltage drop is the measure of the voltage required to cause the\ )Tj 0 -1.375 TD ( current flow through a component.)Tj 0 -2.375 TD (If you have a voltmeter, take some time to do a little playing, which wi\ ll)Tj 0 -1.375 TD ( both make you more familiar with the use of the meter and maybe)Tj T* ( prevent some future electrical problems. Set your meter onto a scale,)Tj T* ( which is just above 12 volts DC, and prepare to do some testing.)Tj 0 -2.375 TD (Refer to your wiring diagram $either from shop manual or downloaded$)Tj 0 -1.375 TD ( and locate the horn circuit. You should be able to trace backwards from\ )Tj T* ( the horn to the horn switch, ignition switch to battery positive. It's \ not)Tj T* ( easy to find and trace the wires but becomes easier with practice. Now)Tj T* ( let's trace the circuit the easy way! Put the negative $black$ voltme\ ter)Tj T* ( lead onto the horn Brown wire terminal and connect the positive $red$\ )Tj T* ( voltmeter lead onto the battery positive. What does the meter read?)Tj T* ( Well, the meter "sees" the voltage difference between the battery)Tj T* ( positive and negative because there is no current flow so no work is)Tj T* ( being done and so there will be no voltage drop between the two points)Tj T* ( being monitored. OK, so now let's do some work! Switch the ignition on)Tj T* ( and hit the horn....the voltage measured now becomes the voltage)Tj T* ( necessary to make current flow through the resistance of the circuit)Tj T* ( between the meter leads. In other words the voltage now displayed is)Tj T* ( the voltage required to make the horn circuit current flow through the)Tj T* ( battery + cable, wires, fuse, ignition switch, horn button and up to $\ but)Tj T* ( not including$ the horn. Record the voltage. 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The more you do, the easier you will find it to be. 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Ohm's Law tells us that one volt is required to forc\ e)Tj ET Q BT 12 0 0 12 141.25 737.25 Tm ( one amp through one ohm of resistance. So the horn 12 volts divided by)Tj T* ( 2.5 amps = 4.8 ohms.)Tj 0 -2.375 TD (Let's say that we have a Vd of 1 volt in the horn positive circuit. That\ )Tj 0 -1.375 TD ( leaves 11 volts, which we will measure across the horn, right? Right.)Tj T* ( Since the horn has 4.8 ohms of resistance with 11 volts, 11 volts divid\ ed)Tj T* ( by 4.8 ohms = 2.9 amps. How much work $honk$ will be done? 2.3)Tj T* ( amps x 11 volts = 25.3 watts, not as loud! Not as much work done.)Tj 0 -2.375 TD (To turn the problem around, the voltage drop will show the work being)Tj 0 -1.375 TD ( done. There should be next to no voltage drop where we want no work)Tj T* ( to be done. Do we want work to be done to force current through the)Tj T* ( horn button? No, we don't. That will simply result in heating of the)Tj T* ( button.)Tj 0 -2.375 TD (What will happen if we install a louder horn? Unless the horn is louder)Tj 0 -1.375 TD ( because it is more efficient then it must be louder because it does mor\ e)Tj T* ( work, has less resistance so more current flow results in more watts of\ )Tj T* ( work at the same voltage. OK but if there is more current flow through)Tj T* ( the same circuit, there will be more voltage drop in that circuit becau\ se)Tj T* ( there is more current flow through the same resistance.)Tj 0 -2.375 TD (Ok, say we install the new horn and it isn't a loud as we hoped? $Is it\ )Tj 0 -1.375 TD ( ever?$ We can do our voltage drop measurements to see how much Vd)Tj T* ( is occurring in the circuits. There should, ideally, be no drop except)Tj T* ( across the horn. If we find a drop elsewhere we can use the voltmeter)Tj T* ( to find it and eliminate it. Voltage drops can be done across parts of \ the)Tj T* ( positive circuit such as across the horn button itself. If we find more\ )Tj T* ( than $as a rule of thumb$ 0.2 volts or more across any wire or)Tj T* ( component $other than the one we want doing the work$ we should find)Tj T* ( a way to eliminate it.)Tj 0 -2.375 TD (One easy way to eliminate the voltage drop across part of the horn)Tj 0 -1.375 TD ( circuit will be to install a relay. The relay will use the horn button \ and)Tj T* ( the other positive side wiring to operate the relay, which requires a v\ ery)Tj T* ( small current, which results in a small voltage drop. This means that f\ ull)Tj T* ( battery voltage will be available to operate the relay. The relay will \ have)Tj T* ( low resistance $low Vd$ in the switched circuit so full battery volta\ ge will)Tj T* ( be available to the horn. This will result in a loud honk and long life\ for)Tj T* ( the horn button.)Tj 0 -2.375 TD (Try similar measures to the headlight, taillight, starter and other circ\ uits.)Tj 0 -1.375 TD ( Any Vd of 0.2 volts or more except across the component intended to do)Tj T* ( the work can be located by moving the voltmeter leads along the circuit\ )Tj ET q 13 36 585 730 re W n BT 12 0 0 12 141.25 38.25 Tm ( in question until we have the Vd between the leads. 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Honk and measure the voltage drop across)Tj T* ( the horn. This is the only useful voltage drop in the circuit but more \ on)Tj T* ( that later.)Tj 0 -2.375 TD (Finally, connect the voltmeter positive $red$ lead to the ground side)Tj 0 -1.375 TD ( $black wire$ connector on the horn and the negative $black$ voltmet\ er)Tj T* ( lead to the ground side of the battery. Honk and measure the voltage)Tj T* ( drop across the ground side of the horn circuit.)Tj 0 -2.375 TD (We have three voltage drops; the Vd in the circuit from battery to horn;\ )Tj 0 -1.375 TD ( Vd across the horn; and Vd in the circuit from horn to battery. The tot\ al)Tj T* ( of the three voltage drops will equal the battery voltage during the)Tj T* ( honk.)Tj 0 -2.375 TD (Now let's consider the meanings of the three voltage drops...which one i\ s)Tj 0 -1.375 TD ( the Vd across the useful work being done? Of course! The Vd across the)Tj T* ( horn is the one indicating the work we want to be done so the others)Tj T* ( must be unwanted. Since we have $in round numbers$ 12 volts available\ )Tj T* ( to push current through our horn, we want to use the whole 12 volts)Tj T* ( across the horn to make the maximum current flow through the horn)Tj T* ( but if any voltage is required to make current flow in any other part o\ f)Tj T* ( the circuit then that much less voltage is available to make current fl\ ow)Tj T* ( through the horn.)Tj 0 -2.375 TD (So what we want to do is to use the measure of voltage drop to find)Tj 0 -1.375 TD ( unwanted voltage drops so that we can decide if they are a problem,)Tj T* ( which needs to be removed.)Tj /TT2 1 Tf 0 -2.375 TD (Some Ohm's Law and Watts:)Tj /TT1 1 Tf T* (The horn is rated at 2.5 amps, 12 volts according to the wiring diagram.\ )Tj 0 -1.375 TD ( What does this translate to?)Tj 0 -2.375 TD (Work being done in an electrical circuit will usually be expressed in)Tj 0 -1.375 TD ( Watts. 60-Watt bulb for example.)Tj 0 -2.375 TD (Watts can be calculated Amps x Volts = Watts. Thus the 60 Watt /12 volt)Tj 0 -1.375 TD ( bulb requires 5 amps while a 60 watt/120 volt bulb requires 0.5 amp.)Tj T* ( Different volts and amps but the same work being done, the same)Tj T* ( watts.)Tj 0 -2.375 TD (We will over simplify by disregarding the dynamic loads but the pattern)Tj 0 -1.375 TD ( will be true.)Tj 0 -2.375 TD (The horn 2.5 amps x 12 volts = 30 watts. OK, but what happens if we)Tj ET q 13 36 585 730 re W n BT 12 0 0 12 141.25 40.5 Tm ( don't see a Vd across the horn of 12 volts? 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