# Ohm's law clarification

I am slightly confused about Ohm's law. I understand V = IR, and I did a simple experiment on paper. The results are slightly confusing, so I was hoping someone could tell me if I'm right?

First, I have an input voltage of 5V at 1A. I then pass it through a 1 Ohm resistor to get 5A at 1V. If I pass it through another 1 Ohm resistor, I get back to my starting voltage and current of 5V at 1A. Is this correct/normal? simulate this circuit – Schematic created using CircuitLab

Edit: Would this work any better? simulate this circuit

• The resistors are in series so the current through them is the same. In this case it is I = 5V/$(R_{1} + R_{2})$ = $5/2$ A.
– Null
Sep 23, 2014 at 21:36
• Don't resistors change the current flowing through them? Sep 23, 2014 at 21:39
• For a given voltage across a given resistance the current is simply I = V/R -- this is simply a restatement of Ohm's Law. By adding the second resistor you have doubled the resistance. Hence I = V/(2R). The current through the resistors has changed but the resistors are in series so the current through them is the same.
– Null
Sep 23, 2014 at 21:43
• This is more than just slight confusion... Sep 23, 2014 at 21:47
• Ah, so I wouldn't use the voltage/current from after the first resistor to calculate the voltage/current after the second? Sep 23, 2014 at 21:48

There are two fundamental laws of electricity that you must always keep in mind - Kirchoff's Current Law, and Kirchoff's Voltage Law.

I consider Kirchoff's Laws as scientific wording of what should be common-sense observations - unfortunately, common sense isn't as common as we might like, so we have to spell these things out...

Kirchoff's Current Law (KCL) states that the algebraic sum of the currents at any point in a circuit must equal zero. In plain English, this means that all the current that arrives at a point in the circuit must leave that point (and no more current can leave than arrives.) If KCL was not true, charges could build up at one point in the circuit, leaving none to flow elsewhere, so a circuit would only operate briefly, until there was no charge available to flow.

KCL means that the current will be the same at all points in a simple series circuit, so your first circuit (with 1 amp at one point, and 5 amps at another) is impossible.

In the first circuit you have a total resistance of 2 Ohms, and (I assume) a power supply of 5 volts. From Ohm's Law, this will result in a current of 2.5 amps, and a voltage drop of 2.5 volts across each resistor.

Kirchoff's Voltage Law states that the algebraic sum of the voltages around a series circuit must equal zero. In simpler terms, applying to your circuit with one power source, this means that the sum of the voltage drops across the resistors in the circuit must equal the supply voltage.

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.

• So, if I was to input 5V into my circuit, I could change the current down but I could not increase the current by adding a resistor. The current is what it is after I input it? Sep 23, 2014 at 21:58
• So if I wanted 10A somewhere in my circuit, I must input 10A? Sep 23, 2014 at 22:01
• Adding resistors in series will increase the overall resistance as seen by the source and decrease the current. Adding resistors in parallel will decrease the overall resistance as seen by the source and will increase the current. Sep 23, 2014 at 22:01
• Two one ohm resistors in PARALLEL will draw 10 A from a 5V source, as two resistors in parallel is 1/(1/1+1/1) = 1/2 ohms, and 5 volts = I * 1/2 ohms makes I = 10 amps. Sep 23, 2014 at 22:03
• Would it work better to make a circuit with 5V->0.5Ohm resistor->10A->5Ohm resistor->1A? Sep 23, 2014 at 22:15

In your first schematic, the nodes are mislabeled -- there is no such thing as a node current. Current only flows in loops. So you can't label a node "5V @ 1A", you can only label the node "5V".

I've redrawn your schematic in conventional form, with the energy flowing from left to right and the higher voltages towards the top. simulate this circuit – Schematic created using CircuitLab

There are three nodes (a node is a "place that has voltage"):

Node voltage V1 is the voltage at BAT1(+) terminal. and one terminal of R1.

Node voltage V2 is the voltage at one terminal of R1 and one terminal of R2.

Node voltage V3 is the voltage at one terminal of R2 and at BAT1(-) terminal.

This is a lumped-constant model, so we're ignoring the minor effects of wiring resistance and inductance, and just assuming that the voltage is the same along the entire length of a wire.

There is one mesh loop (a mesh is a "loop that has current"):

The mesh current i flows through BAT1, R1, and R2. The same current flows through all of these components.

This is a DC circuit because the node voltages and mesh currents are steady, and do not change. For changing signals there are AC circuit analysis techniques. But for DC circuits, the three basic laws that always apply are Ohm's Law, Kirchoff's Current Law (KCL) and Kirchoff's Voltage Law (KVL).

KCL applies at each node, because the total of all currents "in" and all currents "out" of that node, must be equal (i.e. the algebraic sum of all currents going "in" must be zero). This gives one equation for each node.

KVL applies around each mesh, because the algebraic total of all voltages around the loop must equal zero. This gives one equation for each mesh.

Ohm's law applies at each resistor (and this is where the AC analysis gets more complicated). Ohm's law relates the voltage between the two nodes, to the mesh current that flows through the resistor. This gives an equation.

Put all these equations together, do some algebra, and you can determine the voltage at every node and the current through every mesh. For DC circuits analysis, that's how the circuit is "solved".

You are leaking 4 amps worth of electrons somewhere. Nature frowns on this and charge is conserved. Current out of the battery in your closed loop has to equal current in. If you have 5 amps, then the voltage drop across each resistor must be 5V. This means the battery must be a 10V battery.

Or, look at it as a single resistor of 2 ohms. If there are 5 amps in the circuit, then voltage drop is IR = 10V.

Measuring relative to the negative side of the battery, you should get 10V 5V 0V left to right.

• But If I wanted a current of 5A in the middle part, I=V/R, I=5/1, I=5A. What would the voltage be then? I would assume it to be V=IR, V=1*1, V=1v Sep 23, 2014 at 21:44
• And then if I add another resistor, I=V/R, I=1*1, I=1A. Then I would assume voltage to be V=IR, V=5*1, V=5v Sep 23, 2014 at 21:47
• You need to pick something as constant. What is the voltage of your (ideal) battery? Or do you want to set 5 amps as a constant and find the voltage you need? Sep 23, 2014 at 21:52
• I am more concerned with the current, then the voltage. I need the current to start at 1A, go to 3.33A, and back down to 1A. I was thinking the above circuit would do that, with some modifications. The battery I plan to use has 5V output. Sep 23, 2014 at 21:54
• Current is like water in a hose. You can't have 1 gal/minute going in one end, 3 gal/minute at some point along the hose, and back to 1 gal/minute at the end. That would be magic. Sep 23, 2014 at 21:56