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enter image description here

So here's the problem, I made the assumption that D1 and D2 are short, whereas D3 is an open, and combined the 500 ohm resistors in series: enter image description here

But my question is, why is there no Voltage across the 500 ohm resistor that is on the rightmost side? And also no Voltage across the highlighted resistor, isn't it parallel with 1K, and should therefore have the same 20V of voltage?

When I do calculations, there should be voltage there.

So first, you combine the 1k and 500 ohm resistors, which are in parallel. $$\frac{1000 \cdot 500}{1000 + 500} = 333.3 Ω$$

Voltage across parallel resistors is the same, voltage across combined 333.3 Ω is 20V, therefore 20V is also across the 500Ω resistors into which current "I" flows in in the first picture.

"I" current = $$ \frac{20V}{500Ω} = 40 mA $$

40 mA verifies that D1 is indeed a short. Diode conducts, and current is more or equal to 0 across a diode.

Voltage across D3 is (0 - 20V) = -20V

Now to find "V", at the positive side of the 500Ω resistor, it's -20V, at the negative side, it's - 20V because +20V is the voltage across the 500Ω resistors that's kinda between D2 and D3 in first picture.

So when calculating "V" it's (-20V - -20V) = 0V

Correct? But there is in fact voltage across an open, right? Just not across the resistor on the rightmost side.

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  • \$\begingroup\$ um me thinks you are getting your parallels and series confused... try redrawing the circuit properly with the shorts and opens in place. \$\endgroup\$ – Trevor_G Sep 7 '17 at 21:03
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enter image description here

Figure 1. The reds are wrong. The greens are right.

  1. Wrong. It should be 500 Ω, not 1 kΩ.
  2. Wrong. It should be 500 Ω, not a short circuit.
  3. Correct. Replace D2 with a short.
  4. Correct. 500 Ω.
  5. Correct. Reverse diode is open circuit.
  6. Correct. 500 Ω is 500 Ω.

But my question is, why is there no Voltage across the 500 ohm resistor that is on the rightmost side?

Now think! Resistor 6 is part of an open circuit. How much current is flowing through R6? From Ohm's Law what is the voltage across it? (\$ V = IR \$.)


From confused chat:

enter image description here

Figure 2. OP's open-circuit schematic.

Well, there is current from -20V isn't it? Or current won't go from -20V, right? Then it's 0, but(!) AS shown here i.imgur.com/5e6TcLG.jpg, just because current is 0 doesn't mean voltage is 0....

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 3. Fig. 2 redrawn keeping positive at top and negative at bottom.

Figure 3 shows the correct measurement of voltages across the circuit. Since there is an open circuit no current can flow through R1 and R2. Therefore the voltage at both ends is equal so the voltage across each resistor = 0 V. The full 20 V of the two batteries appears across the open circuit.

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  • \$\begingroup\$ for 1, 2. The two 500 Ω are in series, so I combined them into one 1k resistor. \$\endgroup\$ – Jack Sep 24 '17 at 17:16
  • \$\begingroup\$ You can't do that. You would have to take into account the second vertical 500 Ω resistor. This means that you would parallel the two vertical 500 Ω resistors first, giving you 250 Ω and then series that pair with the horizontal (do you see why we always number the components R1, R2, etc.?) 500 Ω to get 750 Ω, not 1 kΩ. \$\endgroup\$ – Transistor Sep 24 '17 at 17:21
  • \$\begingroup\$ Also, regarding your so-called "ohm's law", it's not always true, case in point: i.imgur.com/5e6TcLG.jpg (curren't doesn't flow on either side of Vd2, but there is voltage across Vd2!) \$\endgroup\$ – Jack Sep 24 '17 at 17:35
  • \$\begingroup\$ Jack, I'm not making it up. If you try to apply Ohm's Law (not capital letters) to the open circuit of \$ V_{D2} \$ you have a problem: \$ I \$ is zero and \$ R \$ is infinite. This leads you to \$ V = IR = 0 \times \infty \$ which is not solvable. This is not a defect in Ohm's Law - it's a maths limitation. The answer can be anything. In your case it's 10 V because that's what the voltage source is. \$\endgroup\$ – Transistor Sep 24 '17 at 17:43
  • \$\begingroup\$ also, is R4 rightmost 500 Ω has -20V across it? \$\endgroup\$ – Jack Sep 24 '17 at 17:48
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The first thing I do, if a circuit isn't immediately obvious to me, is to work on redrawing it.

  1. Current flow should be arranged so that the top of a schematic is the most positive and the bottom of the schematic is the most negative. Signal, if applicable, should flow from left to right, with inputs on the left and outputs on the right.
  2. Don't bus voltage rails (or ground) around. You don't need to see all the connections as it doesn't matter (mostly) for understanding a circuit. And just label the nodes where you know the voltage.

This will save you a lot of grief.

schematic

simulate this circuit – Schematic created using CircuitLab

It's pretty easy to see now that \$V_X\$ will be somewhere between \$+20\:\textrm{V}\$ and \$0\:\textrm{V}\$ and that \$D_1\$ is on. You can also just imagine that all of the circuit inside the green box is removed for a moment and this also allows you to see that \$D_1\$ is definitely forward-biased and therefore on. This means that, indeed, \$V_X\$ will be somewhere between \$+20\:\textrm{V}\$ and \$0\:\textrm{V}\$.

Given that, let's add in the circuit inside the green box but still excluding the circuit inside the blue box. Knowing that \$V_X\$ will be somewhere between \$+20\:\textrm{V}\$ and \$0\:\textrm{V}\$, tells us that it is at least "positive-valued" and therefore it is easy to see that \$D_2\$ is also forward-biased and therefore on, too.

Given that, it's easy to see that since \$D_2\$ is an ideal diode and on that \$V_Y=V_X\$ and is also positive. This clearly means that \$D_3\$ is reverse-biased and therefore off.

This confirms your assumptions.

Now just replace all those ideal diodes with an appropriate "short" or "open", as determined earlier:

schematic

simulate this circuit

The rest is pretty easy from there.

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