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<blockquote data-quote="jrz126" data-source="post: 1337548" data-attributes="member: 556466">It&#039;s not the test leads that would fail, but the unit itself. (well the leads would fail too if you tried measuring the current from the starter)...but anyhoo, the way those amp meters work is by placing a resistor with a very low resistance (lets say 0.1 ohm) in series with the circuit to be measured. Then the meter would measure the voltage drop across that resistor. Then using good old ohms law, the current is = to (voltage across resistor)/(resistance).Now this becomes a problem when you are trying to measure high currents in 2 ways: 1. the voltage drop created by the resistor in the meter. (say you are measuring a 12V, 10amp unit with the meter, there will be a 1v drop across the meter, giving you only 11V to the unit.) and 2. the power disipated by it. power=I^2*R (at 10 amps and .1 ohms, you&#039;ll need a 10 watt resistor inside the meter).Hmmm...I could talk about this stuff for hours...Oh yeah and Flukes &amp;gt; all</blockquote>

[QUOTE="jrz126, post: 1337548, member: 556466"] It's not the test leads that would fail, but the unit itself. (well the leads would fail too if you tried measuring the current from the starter)...but anyhoo, the way those amp meters work is by placing a resistor with a very low resistance (lets say 0.1 ohm) in series with the circuit to be measured. Then the meter would measure the voltage drop across that resistor. Then using good old ohms law, the current is = to (voltage across resistor)/(resistance). Now this becomes a problem when you are trying to measure high currents in 2 ways: 1. the voltage drop created by the resistor in the meter. (say you are measuring a 12V, 10amp unit with the meter, there will be a 1v drop across the meter, giving you only 11V to the unit.) and 2. the power disipated by it. power=I^2*R (at 10 amps and .1 ohms, you'll need a 10 watt resistor inside the meter). Hmmm...I could talk about this stuff for hours... Oh yeah and Flukes > all [/QUOTE]

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