Extremly basic electronics question

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Original post by Witchcraven I have not really learned much of electronics. I have built rather complex digital circuits, but it is probably lucky I got the voltages right and didnt blow something up (except my power supply once). So this semester I signed up for an electronics class. We have not gotten far at all yet, but I have been playing around a little in circuit simulators. In the sim I built a very simple circuit that contains a 10v dc supply and 3 100 ohm resistors in series. As is my understanding of series circuits, the voltage across each resistor should be 3.33volts. The simulator confirms this. However when I look at the voltages on the connection wires, the simulator says 10v, 6.67v, 3.33v, and 0v. But that is not really quite accurate is it? If I replaced the wire that reads 6.67 volts in the sim with a component (such as a lamp or something) the component would not recieve 6.67 volts. It would be based on the internal resistance of the component, just like a resistor, correct? Is it just notation the sim uses in case I wanted to bring wires off in parallel to make a voltage divider?

The simulator is checking the voltage at the node under the assumption that it is relative to ground. So just after the battery, relative to ground, you are checking the voltage drop after all three resistors, 3.33 * 3 = 10. After the first resistor, you're measuring the drop across the last two. 3.33 * 2 = 6.67. And so forth.

*edit: just to be perfectly clear, the simulator is not measuring the voltage drop across the wire connectors. That should be zero, always. It is measuring the voltage drop between your two probes. When you check the resistor, you get probes on either side. When you check the wire, you get one probe at the wire and one probe at ground.

CM

I see. But perhaps my confusion comes from why they call it voltage drop. As in series, its not like the voltage after the resistor is lower because of it right? Because the voltage across a component in a series circuit is based on its resistance.

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Original post by Witchcraven I see. But perhaps my confusion comes from why they call it voltage drop. As in series, its not like the voltage after the resistor is lower because of it right? Because the voltage across a component in a series circuit is based on its resistance.

Relative to ground, the voltage after the resistor is lower than before it. That is what you measure when you check the voltage of the resistor, the difference in voltage on one side relative to the other.

Think of resistors like a series of steps. You have 10 feet to go in three equal steps. If you measure the height of the first step, you get 3.33 feet. But if you measure the distance from the ground to the bottom of the first step, you get 6.67 feet.

CM

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Original post by Witchcraven I see. But perhaps my confusion comes from why they call it voltage drop. As in series, its not like the voltage after the resistor is lower because of it right? Because the voltage across a component in a series circuit is based on its resistance.

Consider that voltage is entirely *potential*, that is, the potential of the electrons to supply a certain amount of energy. It also cannot be measured directly; the only "voltage" that matters is the difference in voltage between two nodes in a circuit.

You have a 10v power supply, meaning that the voltage will rise 10 volts from the negative terminal to the positive. Because of kirchoff's law, the sum of all voltage drops around a loop must be 0. Because the power supply is a "+10V" voltage increase, the voltage drop over the rest of the components must be 10V, or a -10V "increase" If you have 3 equal resistors in series each gets a drop of 10/3V, or 3.33V.

Now look at the current picture, which is absolute(i.e., no need for "relative" currents right now) Some amount of current is flowing through the circuit: because in a series circuit the current is constant, according to V=IR, since the sum of voltage drops = 10V, and the total resistance is equal to the sum of the resistances, we can solve for the current passing through the circuit.

Now we know the current passing through each resistor, because it is the same as for the whole circuit. According to V=IR the voltage drop is 1/3 the total voltage drop (so the drop is 3.33 volts)

For the first resistor in series, the current entering the resistor has 3.33V more potential than after. The resistor doesn't care about absolute voltage, only the voltage drop between the node before the resistor and after matters.

So if the entering voltage is 10V (relative to the -ve terminal of the power supply) and the resistor drops 3.33V (absolute, since it's a drop) then the voltage at the node exiting the resistor is 6.667V(relative to the -ve terminal of the battery)

Gotcha. So then the ways I could place a component that needs, say 3.3volts on a 10 volt circuit would be either to figure out its internal resistance and put other resistors on the circuit so that it draws 3.3volts, or go parallel over the resistor voltage drops. I suppose though going parallel would need some more considerations as resistance in a parallel network is different.

Consider this:

_________________________________ 10V    |    |   ---   | | 1K   | |   ---       __________    |       |          |(A) +-------|   COMP   |--------- Output    |       |__________|       ---            |   | | 1K         |   | |            |   ---            |    |             |    |             |____|_____________|______________ GND

If the component COMP wasn't connected you'd have 5V (relative to ground) at the node (A) between the 2 resistors. If COMP has a relatively large internal resistance like 100K (and thus draws a relatively small amount of current), the effect it will have on the voltage at node A will be negligable so it can be connected as shown without having to consider its effects.

To work out the actual voltage at A, consider the bottom 1K resistor and COMP to be in parallel. The equivalent resistence is 1/((1/1K)+(1/100K)) = 100K/101, which is only slightly under 1K, which would push the voltage at A slightly under 5V, accurate to 3 digits its 4.975V.

However, if COMP's resistance isn't relatively high then it can have a significant effect on the voltage at A. Lets say the resistance of COMP is 1K, so the equivalent resistance is 1/((1/1K)+(1/1K)) = 0.5K. This is a huge change of resistance and would case the voltage at A to drop to 3.33V, which is definately significant.

Rule of thumb, if the amount of current you draw doesn't matter just make sure your resistors in the voltage divider are significantly smaller than the internal resistance of the component. If the current you draw does matter....I hope you enjoy math [smile]

NB: In electroncs, significantly larger (symbol for which is >>) is usually consider at least 10x