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As you can see from the picture, shouldn't the 100 Ohms be the only current influencing factor at whatever the voltage it is supplying? How does Potentiometer on the right side cause decrease in current when I am measuring the current even before current crosses potentiometer?

For example, at 4 V, 100 Ohm(fixed) on the left / 1000 Ohm(potentiometer) on the right I measured 3.59mA

Then at 4 V, 100 Ohm(fixed) on the left / 2000 Ohm(potentiometer) on the right I measured 1.88mA

Am I wrong? Clearly experiment say so...

Potentiometer

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In a simple series circuit, the voltage is proportionally divided by any elements in the circuit, while the current is the same at all nodes. In a series circuit, you will not see different currents before or after an element (in this case the various resistors).

In a simple parallel circuit, the current is divided by the parallel nodes.

It's not like water, where a large hole (the 100 ohm resistor) will allow a lot of water through through it, and then a small hole (the pot) will only allow a small part of that water though it.

The Pot essentially sets the current through the entire circuit, even though it's after your measuring point.

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  • \$\begingroup\$ It's exactly like water, where the battery voltage is like the pressure exerted on the water main at the bottom of the column of water under a water tower, the diameter and length of the water main is the resistance of the 100 ohm resistor, and the diameter and length of the service leading into your home (with a faucet on the end of it) is like the pot. \$\endgroup\$ – EM Fields Mar 18 '16 at 8:14
  • \$\begingroup\$ @EMFields that... just makes it more complicated. And a pressure reduction like that, in water, would result in OP's assumption, where measuring the pressure at node A, the 100 Ohm, would be higher, then it would at node B, the pot. A strong river isn't stopped by the weak flow at any point downstream. \$\endgroup\$ – Passerby Mar 18 '16 at 14:53
  • \$\begingroup\$ With no water flowing, the pressure in the water system piping is everywhere the same, as is the voltage in the electric system with no current flow. However, once flow exists, there will be a pressure drop across the large pipe (the 100 ohm resistor) and the small pipe (the wiring leading up to the pot) which will limit the flow through the faucet (the pot). Looking at it another way, consider the 100 ohm resistor to be a partially open dam across a strong river and it becomes apparent that there will be a pressure drop across the dam (a flow limit) which can't be recovered later. \$\endgroup\$ – EM Fields Mar 18 '16 at 16:09
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The current in the circuit will be determined by the total resistance in the circuit. If the potentiometer is set to 1000 ohms, the total resistance will be 1100 ohms, so, with a 4 volt power supply, the current will be 3.6 mA (I = E/R = 4/1100). If the pot is set to 2000 ohms, the total resistance will be 2100 ohms, so the current will be 1.9 mA. (Your numbers may vary a bit, depending on how accurately you set the pot, and on the resistance of the ammeter).

Kirchoff's Current Law says that in a simple series circuit such as this, the current is the same at all points in the circuit.

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You're not wrong. :)

The total resistance of resistors in series is the sum of the individual resistances, so in your first example that total resistance would be:

\$ Rt = 100\Omega + 1000\Omega = 1100\text { ohms}\$.

From Ohm's law, we have:

$$ I= \frac{E}{R}$$

So, knowing that E equals 4 volts and R = 1100 ohms, we can solve for the current through the string like this:

$$ I = \frac{4V}{1100\Omega} \approx 3.64 \text{ milliamperes} $$

And, in your second example with a total resistance of 2100 ohms, we have:

$$ I = \frac{4V}{2100\Omega} \approx 1.9 \text{ milliamperes} $$

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