When I connect a multi meter across BC and AC I get the same reading, both show 5V, but isn't the resistor supposed to decrease the current/voltage?
simulate this circuit – Schematic created using CircuitLab
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7\$\begingroup\$ Only if there is a current through it does it drop voltage. \$\endgroup\$– periblepsisSep 24 at 16:15
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\$\begingroup\$ there is current through it, multi meter make it a full connection. \$\endgroup\$– DanSep 24 at 16:18
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\$\begingroup\$ That current is very small. Not enough for you to notice a difference. Try making R1 a thousand times larger. Or ten thousand times larger. \$\endgroup\$– periblepsisSep 24 at 16:20
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\$\begingroup\$ Try a 10Meg resistor and you'll see some voltage drop. \$\endgroup\$– user1850479Sep 24 at 16:22
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11\$\begingroup\$ The resistor in the diagram is not across the 5V source as indicated in the title. It is in series with the source. \$\endgroup\$– RussellHSep 24 at 16:27
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3 Answers
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Many multimeter manufacturers have standardized their equivalent input resistance at ten MEGohms. Some precision voltmeters have much higher input resistance. But let's assume your multimeter looks like a 10M resistor:
simulate this circuit – Schematic created using CircuitLab
You have a voltage divider, which means that \$V_m < 5V\$. But not much less. \$V_m\$ is what the multimeter will display.
You might use a 10V or 20V scale to display the 5V result: a 2V scale would over-range:
- A 3.5 digit multimeter would display 4.99, or 5.00
- a 4.5 digit multimeter would display 4.999 or 5.000
Since \$V_m = 4.9995\$ volts, the display resolution is too coarse to resolve the difference between 5.00000 and 4.9995
If your multimeter had better resolution, you should be able to see the drop.
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\$\begingroup\$ Ok but I still don't get why the the voltage doesn't drop significantly until I add something else to the circuit like and LED? when I add an LED voltage drop to around 2 V. V = IR, in my circuit 5 = I * 1000, then I is 0.0005 amps which means there is barely anything going through the resistor itself? how is this possible. \$\endgroup\$– DanSep 24 at 22:53
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\$\begingroup\$ Oh is it because the multi meter has such a strong internal resistor that barely any current goes through it? \$\endgroup\$– DanSep 24 at 22:55
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5\$\begingroup\$ @Dan yes. The resistance around the loop is 10,001,000 ohms and the current is just under half a microamp. \$\endgroup\$– hobbsSep 25 at 0:40
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\$\begingroup\$ @12431234123412341234123 Collins dictionary verifies that megohm is one million ohms, so 10 megohms is ten million ohms. But I note that peak usage occurred in 1948, falling since then. Have I sorta dated myself? Some folks here get M confused with m, so I thought that MEG makes a better distinction. \$\endgroup\$ Sep 25 at 16:19
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\$\begingroup\$ @glen_geek The SI brochure is very clear that M is Mega for 1'000'000 and m is milli for 0.001 . I didn't find any occurence of MEG, Meg, MEGohm, Megohm, MEGohms nor Megohms in there. If you want to be clear, just use 10 MΩ, 10e6 Ω, 10'000'000 Ω, 1e6 ohm or 10'000'000 ohm. And why the s at the end? There is no second in this unit. \$\endgroup\$ Sep 26 at 7:25
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The voltage dropped across the resistor is proportional to the current flowing through it; the constant of proportionality is the resistor's value: V = I*R. When the resistor does not have current through it, as when it is open circuited or connected to a multimeter which has an impedance orders of magnitude higher than 5k ohms, there will be no measurable voltage drop across it.
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Try to see the picture as a liquid pipe with a reducer (resistor). As long as the reducer is blanked the pressure (voltage) will be the same before and after it. If you leave the liquid flow (current) pass then the outlet pressure will be less than the inlet.
