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Okay, this time we're going to check for current. I have the same setup as I did for when we were checking for voltage in this circuit
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I have the same simple series circuit here. I have our positive power supply coming in, going to this orange jumper wire. The orange jumper wire goes to the center pin of the slide switch
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And the far right pin of the slide switch is in connection with the anode of the blue LED
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and then the cathode of the LED is in line with this 1,000-on resistor
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and the other end of the leg of the resistor is in line with the negative supply here
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So in order to check for current, I need to remove this LED out from the circuit
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and all I'm going to do is I'm going to turn it and put the anode of the LED all of the LED all
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off on this point here and I'm going to keep its cathode connected to this resistor here
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So right now the LED is out of the circuit. The circuit is open between the switch and the LED here
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And what I'm going to do is I'm going to put a jumper wire here so I can attach my probe from the multimeter
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I'm putting it here because that's where I took out. the anode of the LED from the circuit
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So now in order for me to make a connection with one of my probes from the multimeter
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I needed to put this jumper wire here. So now I'm going to take my red lead of my
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my multimeter, my test probe to this orange jumper wire. And I'm going to take the black lead of my multimeter
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and I going to attach its probe to the anode of the LED So right now I got the LED out of the circuit and in between where the circuit is opened
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between the, uh, oops, between the far right pin of the switch
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and the anode of the LED, the circuit is open, and I have my
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test probes from the multimeter connected in between that open of the circuit
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So now I need to set my multimeter over to DC amps and this symbol here, the straight line
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next to this capital A means that this is to measure amp-age in the circuit for DC circuit
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And I'm going to put it to 200 milli-amps, the 200m here
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So I'm going to turn from off to 200 millie amps right here
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I'm going to turn on the circuit now, so I'm going to turn the switch on
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You can see the LED is lit up. And I'm getting a reading of 6.4 milliamps
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So if I turn my calculator on and we look back on to Oams Law
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6.4 millie amps is written 0.0064 amps
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So 6.4 millie amps is the same as 0.0064 amps. I'm just changing the unit size
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And if we remember from Omslaw, v equals IR, so I is the current, and that's the current that we're reading
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the 6 milliamps or amps and I multiply by the resistance in the circuit which we know is this 1 resistor and I hit enter I get 6 volts
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So remember, a series circuit is a voltage divider. So the 6.4 volts is a voltage that's being dropped
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across the resistor. Now if we turn this off and I turn the circuit off and I remove my
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probes and I put the LED back in the circuit again and I'm going to turn the switch
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back on. The LED is on. I'm going to place my black probe which is to my
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common of my multimeter, I'm going to place the red probe on the cath, the anode of the
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LED, and I'm going to turn, I want to measure voltage across the LED right now
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So I set it to 20 volts because this is a 9 volt battery, just like we did when we were checking
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voltage, I needed to set it to a higher voltage than the power source and in this case I sent
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by a multimeter to 20 volts. I'm getting a reading of 2.75 or 2.76 volts
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I'm going to use 2.76 volts, 2.76. I've entered that number into my calculator
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And if I add the voltage that we received by doing the calculation from the current that we
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measured and the resistance that we have in the circuit, this 1,000-on resistor
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If I add these two together the 2 volts I reading across the voltage drop across the LED plus the 6 volts we measured earlier That 9 volts
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So if I remove the probes from the multimeter, and just like I did when measuring voltage
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if I place the orange probe on the positive side, and I place
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the black probe on the negative side and I place my black probe from the
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multimeter on the black jumper wire and I place my red probe from the
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multimeter onto the orange jumper wire. I get 9.05 volts about that which is it's a
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little off but it's very close. But it's pretty much on par
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We're only off about one-tenth of a volt here, which is not a big deal
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It's basically about the same. We're using a 9-volt battery. We get 9 volts
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So we can see that how we check the current, we can use the current reading that we have
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and then the resistance that we know in this circuit. We multiplied those two values together using only
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law to get the voltage. And we measured the voltage drop across the blue LED, which
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gave us a voltage drop of 2.76 volts. And recalling that a series circuit is a voltage
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divider, we have a voltage division across the blue LED and a voltage division across
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the resistor, since these are the only two components in this circuit. And we add those
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two voltage drops in our circuit together and we should get the voltage supply
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which is about 9 volts
