batteriescurrent measurementohms-lawresistancevoltage
simulate this circuit – Schematic created using CircuitLab
Connecting various resistors to the same two batteries rated 1.5 V and 9 V as shown in schematic above returns unexpected results: lower the resistance of R1, lower the current in respect to expected value (as per Ohm's Law).
Would appreciate explanation or hint as to why this happens.
They are simulating a (more or less) real 9V battery. They've modeled the battery as an ideal voltage source of 9V with a series resistance of ~1.5 ohms.
simulate this circuit – Schematic created using CircuitLab
If you work this out, the current is 9V/(11.5 ohms) = 0.783A, so the voltage across the 10 ohm resistor must be 7.83V.
You can't specify the current AND the voltage. Either you are applying 5V or you are applying 1A. Since you have a batery symbol drawn, I will assume you are applying 5 volts.
This 5v is applied across two 1 ohm resistors in series. Total resistance of two 1 ohm resistors in series is 1 + 1 = 2 ohms. V = I * R tells us that 5 = I * 2 where I = 2.5 A. Then the voltage across each resistor is V = 2.5 * 1 = 2.5 volts.
How about applying 1 A to two series 1 ohm resistors? Well, that 1 A is going to produce V = 1 A * 1 ohm = 1 volt across each resistor. Since there are two 1 ohm resistors in series, the voltage across the pair is 1 + 1 = 2 volts.
The current must be the same at all points along that path as charges cannot be created or destroyed ('what goes in must come out'). The voltages around the loop must also add up to zero ('what goes up must come down'). In this case, you go up 5 volts in the battery, then you come down 2.5 volts in each resistor, ending up at zero right where you started.
Best Answer
Your schematic is wrong. The correct one is this:
simulate this circuit – Schematic created using CircuitLab
The Amperemeter has an internal resistance Ramp of maybe 0.1Ω, to measure a voltage over it (that voltage is then shown as "Ampere" I=Vamp/Ramp), and more important, the battery also has an internal resistance, too. The 9V block in particular is made of six very small 1.5V cells, and their already high internal resistances add up due to the series connection.
The lower the resistance Rtest is, the more these low internal resistances matter.