# Solving a circuit with a directional loop

I came up with the following circuit:

Please disregard the labeled values in the diagram - the resistances are identical for the sake of simplicity, and of value $R$, and the voltage has value $V$. Assume ideal circumstances regarding the diode, wires, etc.

I'm not able to solve it because I can't seem to find the total resistance.

As stated in the diagram, there is a parallel branch which splits into a diode facing against the positive end of the circuit and a resistor, in series with an identical resistance.

In my attempt at solving this, I imagined that once the positive current reaches the bottom node, it splits into current going through the resistor towards the negative terminal and current that re-loops through the resistor that was just passed.

Given that assumption, the circuit acts as a series parallel circuit whose net resistance can be found by $$\Sigma R = R + x$$ where $x$ is the value of the resistance in the infinite parallel branches that result from the current re-looping. Then, $x$ would seem to be given by $$x = (\frac{1}{R} + (\frac{1}{R} + (\frac{1}{R} + ...)^{-1})^{-1})^{-1}$$ $$x = (\frac{1}{R} + x)^{-1}$$ $$x^2 + x/R - 1 = 0$$ $$x = \frac{-1/R + \sqrt{1/R^2+4}}{2}$$ But this value of $x$ isn't even dimensionally correct!

Is there another approach I can use to find the resistance?

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In my attempt at solving this, I imagined that once the positive current reaches the bottom node, it splits into current going through the resistor towards the negative terminal and current that re-loops through the resistor that was just passed.

It might do that. But you know that both the resistor and the diode are passive devices. And since they're connected in parallel, the voltage across them is the same. Since they're passive devices (meaning, they're not adding energy to the circuit like a source can), current can only flow through them from high voltage to low voltage. This means you know that current through the diode, if it goes at all, has to go in the same direction as it goes through the resistor.

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But, in order for the diode to have some sort of potential difference going the other way, you would need it to have some nonzero resistance, right? But then this falls apart, since it contradicts with the initial assumptions, that our diode is ideal? – VF1 Feb 20 '13 at 18:38
@VF1 As an ideal diode, its resistance is infinite, i.e., it looks like an open circuit. – Alejandro Feb 20 '13 at 18:48
@Alejandro, but only in one direction, right? – VF1 Feb 20 '13 at 18:48
@VF1, no. The diode doesn't have a potential difference going "the other way" (I assume you mean, higher at the bottom, lower at the top) because it's in parallel with the resistor and we already established that the resistor's potential is higher at the top and lower at the bottom. Parallel connection means the potential difference is the same between the two parts. – The Photon Feb 20 '13 at 18:49
@ThePhoton So because of that, it just gains effective resistance the other way? – VF1 Feb 20 '13 at 18:53

The diode is reverse biased, so no current will flow thru it (other than leagkage, but you said ideal components and that would be much much smaller than the resistor currents anyway). You therefore are simply left with two 4.7 kΩ resistor in series with a 5 V supply. The resistances add, for a total of 9.4 kΩ. From Ohm's law, the current is 5V / 94.kΩ = 532µA. Since the resistors are equal, the node between them will be at half the supply voltage, which is 2.5 V. Yes, it really is that simple.

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But even if the diode is reverse biased, what prevents current from flowing through the other way? – VF1 Feb 20 '13 at 18:40
@VF1: The fact that it is reverse biased. This is how diodes work. When you apply a voltage accross them in reverse, current doesn't flow. That is in part what makes a diode a diode. A resistor, for example, is different. Current flows thru it proportional to the voltage accross it. – Olin Lathrop Feb 20 '13 at 18:53
@VF1, If the diode is reversed biased, and current were flowing through the other way, that would mean the diode is delivering power to the rest of the circuits. And diodes simply can't do that. – The Photon Feb 20 '13 at 18:53
@VF1: Think of the diode as a one-way valve in a water pipe. If there is 2.5 PSI backwards accross the valve, it will block current. Water can't come around somehow and flow forwards thru it. – Olin Lathrop Feb 20 '13 at 18:57
@OlinLathrop It makes sense with the water analogy. – VF1 Feb 20 '13 at 18:59

If you want to solve this problem by hand, and not for example via the use of numerical methods on a computer, you will need to make some assumptions on the diode. Specifically, you will need to decide how sophisticated the model of the diode will be.

For example, two common models are:

1: The ideal diode:

Which has infinite resistance when OFF (open circuit), and null resistance when ON (short circuit).

2: Voltage source and limiting resistor model (more accurate): Where the diode on the model is ideal.

Solution
Since you are considering the diode ideal, then you have to solve this in two steps.

• Suppose that the diode is ON
Then, you have a short circuit and therefore the parallel resistor is totally ignored. In that case the the resistance is simply 4.7 kΩ. But look at the graph on the first link. The current has to be POSITIVE, but in this case it's NEGATIVE.

This supposition is false, the diode isn't ON.

• Suppose that the diode is OFF
Then, you have an open circuit and therefore two resistors in series. In that case the resistance is 9.4 kΩ. But, does this actually make sense? Note that since we have two equivalent resistors, then the voltage across the parallel diode is -2.5V (notice negative sign), which satisfies the graph on the first link.

This supposition is true.

Therefore the answer (finally) to your question, using the ideal diode model, is 9.4 kΩ.

EDIT -
I noticed the poster has been having doubts about the possibility of having a looped current.
Well let's consider that possibility. We can at least agree that the current is in the direction to where the diode is pointing. In that case, if the current $i_D$ loops to the parallel resistor there will automatically be a voltage across that same resistor according to: $$V = R i_D$$

Since the diode is parallel to that resistor, then it also is subjected to that voltage. But that means it would have a negative voltage, making it impossible to be conducting current.

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No, the diode is off. – Olin Lathrop Feb 20 '13 at 21:28
I got confused by the polarization of the source (switched it around). Post edited. – jhc Feb 20 '13 at 21:31

In this case the diode is open, since the 5V supply is generating a current in the direction of the diode's cathode. Since the diode is open, the resistance are connected in series and you can add the values.

In general, as a first approximation (ideal diodes), you can assume that diodes are either open or conducting. You can then assume one of these two states and solve the circuit. Then you test your assumption. If you assumed that the diode was open, then you need to check that the voltage between the anode and the cathode is negative, if it is not, then your assumption is wrong. If you assumed that the diode is conducting, then you need to check that the current (defined as positive when entering by the anode) is positive. Again, if it is not, then your assumption is wrong.

In this example, if you assume that the diode is conducting, then the current will be negative, implying that the assumption is incorrect. If you assume that the diode is open, then the voltage between the anode and the cathode is -2.5V, implying that the assumption is correct.

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But how does that explain the potential for the current to "loop around?" Why wouldn't it? – VF1 Feb 20 '13 at 18:23
@VF1 What do you mean by "loop around"? Note that since the diode is open, the current through it is equal to 0. – Alejandro Feb 20 '13 at 18:29