Why does point B have a voltage of 3.33 V? I was studying diodes in the Sedra book and I came up with the following example; the solution that he provides is the following:

$$ V_b = -10 + 10\times1.33 = 3.33~\mathrm{V} $$ I know why the current has a value of 1.33 A, but how did he come up with this voltage value is a mistery to me. What technique did he use?

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  • \$\begingroup\$ -10 + 10 × 1.33 is equal to 3.3, not 3.33 \$\endgroup\$
    – jsotola
    Commented Jul 15, 2023 at 17:05

2 Answers 2


I know why the current has the value of 1,33A

It is not. It's 1.33 mA.

how he came up with this voltage value is a mistery for me. What technique he used?

Knowing the current, you now can find the voltage across either of the resistors. Then add it or subtract it from one of the supply voltages to get the voltage at node B.

The formula you gave is calculated from the -10 V supply, plus the voltage across the lower resistor, to get the voltage at B.

  • \$\begingroup\$ I see, regarding the 1,33A and 1,33mA it was a mistake. But basically what was used was ohm's law? \$\endgroup\$
    – Matheus
    Commented Jul 15, 2023 at 17:01
  • \$\begingroup\$ Ohm's law and an understanding of potential differences. \$\endgroup\$
    – The Photon
    Commented Jul 15, 2023 at 17:02

Is that who the formula was actually presented? It's numerically correct, but missing the proper units.

Basically, the first thing that was done was calculating the current through both resistors (20V / 15k amounts to 1.33mA, not 1.33A). Then the voltage drop across the 10k resistor was calculated with 1.33mA * 10k = 13.33V. So the voltage at point B is 13.33V above -10V or 3.33V above 0V, which is the final answer.


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