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I wanted to know what the rules are, if a resistor connects to parallel resistors(The R3 resistor), how the current flows, and what determines that. The question was find the total resistance of the network, and the current going through each resistor. We solved this in my class setting up system of equations, like the current going through R2 plus current going through R1 should be equal to initial current, and Kirchhoff's Voltage Law(Voltage loops). The answer we got is 10A(total current), but the teacher did not know the solution off hand. He also assumed that current would flow left to right, because the left resistor in the initial parallel had less resistance so more current. He said this would control the flow through the wire, but I thought voltage would control which way the current flows. And since they are parallel, the voltage would be the same, so no current would flow in the R3 resistor. Any explanation to the problem would be appreciated, specifically what are the rules about current through the R3 resistor?

R3 is the resistor connecting the two. 14v battery

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    \$\begingroup\$ Do you know how to "Thevenize" a voltage divider? (To convert a voltage source and two series resistors into an equivalent, but different, voltage source and a single, but also different, series resistor.) If so, you can Thevenize \$R_1\$ and \$R_4\$ with \$V\$ and then also, separately, Thevenize \$R_2\$ and \$R_5\$ with \$V\$. Then you can easily work out the current in \$R_3\$. \$\endgroup\$
    – jonk
    Apr 30 at 21:49
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You can tell which direction the current will flow by removing the resistor. The left side will have (2/3) of 14V and the right (1/3) of 14V so the current will flow left to right (of course the voltages will change when the resistor is replaced.

You can also solve this for the actual current by finding the Thevenin equivalent of each side and then calculating the current.

It "looks like" a source of (1/3)14V with a source resistance of 2(1||2) = 4/3 ohms. Plus the 1 ohm R3 gives us 7/3 ohms, so the current through R3 can be calculated.

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  • \$\begingroup\$ 1 and 2 ohms in parallel is 2/3 ohm on each side. Thevenin voltages are 1/3 and 2/3 of 14V. Connect all that up in series with and R3 and its current is ... exercise left to the student. R3 IR drop is then known, so junction voltages are known, solve for currents, sum, calculate total resistance. \$\endgroup\$ Apr 30 at 22:15
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The first step in solving for current in a circuit is to set up a sign convention for currents. One does this by drawing arrows along every branch of the circuit. What the arrow means is that when you speak of a positive current through that branch, then the current is going in the direction of the arrow. It doesn't matter which way the arrows go. If you choose an arrow going "left", and the current actually flows "right", then you will just get a negative value for the current. That's all.

So, when your prof

assumed that current would flow left to right

That was fine. However, the rationale

because the left resistor in the initial parallel had less resistance so more current. He said this would control the flow through the wire,

is misleading. One doesn't need a rationale to choose a sign convention.

Now, a point of terminology. You wrote:

if a resistor connects to parallel resistors(The R3 resistor),

There are no parallel resistors in this circuit. For two resistors to be in parallel, both ends of the resistors have to be connected together. No pair of resistors in the circuit matches that criterion.

I thought voltage would control which way the current flows.

It will.

And since they are parallel, the voltage would be the same, so no current would flow in the R3 resistor.

Again, there are no resistors in parallel. However, there are two voltage dividers that are of special interest. The first consists of R1 and R4, and the second consists of R2 and R5. If these voltage dividers did have equal resistance ratios, then you would be right. There would be no voltage across R3. But the resistance ratios are different. R1/R4 = 1/2, while R2/R5 = 2/1 = 2. So, these voltage dividers do not provide equal voltages across R3, and so current will flow through R3.

There are a number of techniques to find the current in circuits such as the one you have. I imagine you set up a set of simultaneous linear equations using Kirchhoff's laws. An alternative, is to use series, parallel, and delta-wye transformations and iteratively simplify the circuit.

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  • \$\begingroup\$ I thought that this was parallel, because of the of node creating R1 and R2. But by your definition, these would not be in parallel. He hasn't taught us what a voltage divider is, he just wanted to cover the basics of electricity in a high school course. Also the current flowing left to right meant that current would flow left to right across the R3 resistor. \$\endgroup\$
    – Max K
    Apr 30 at 22:32
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but I thought voltage would control which way the current flows.

It does.

And since they are parallel, the voltage would be the same, so no current would flow in the R3 resistor.

Which voltage? The voltage at the node where R1, R3 and R4 come together will be higher than the voltage at the node where R2, R3 and R5 come together. So current will flow in R3.

You can figure this out qualitatively by noting that R1 and R4 form a voltage divider, and if R3 were removed from the circuit, the voltage at that node would be \$\frac{2}{3} 14\mathrm V\$. By the same token, if R3 were removed, the voltage at the common node between R2 and R5 would be \$\frac{1}{3} 14\mathrm V\$. So when you connect R3, current must flow.

Note that if you're clever and determined, and if you understand how Thevenin equivalent circuits work, you could work this out without explicitly using KVL or KCL. It'd probably be more work in the end, and there'd either be be some intuitive hand-waving involved that wouldn't get past the sticklers, or so much work proving that you're right that it'll really be more work than just using KVL or KCL.

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  • \$\begingroup\$ I'm not sure I understand why when you remove the R3, one of the voltages would be higher. I'm assuming that when you remove R3, you get parallel series. So when you do this, the voltage in each wire/resistor would be the same. Would this be the case if instead of removing R3, you removed the whole wire connecting the two, because then there is no voltage drop? \$\endgroup\$
    – Max K
    Apr 30 at 22:51
  • \$\begingroup\$ Why don't you do the math. You've got R1 and R4, with the bigger resistor at the low end; then you've got R2 and R5, with the smaller resistor at the low end. If the voltage difference doesn't leap out at you do the math and report back. \$\endgroup\$
    – TimWescott
    Apr 30 at 23:02

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