Ohm's law clarification

Question

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}

Answer 1

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.

Answer 2

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.

Answer 3

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".

Answer 4

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.