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I'm trying to find an expression for the current through the resistor \$R\$ when the AC-source is at max voltage of 5 volts.

\$U_T\$ represents the voltage drop across each diode.

\$R_D\$ represents the the internal resistance of each diode.

As you can see on the picture, I'm wondering how the expression for the current \$I\$ through the resistor is derived?

Now I got the wrong answer if I said that after all the voltage drops the voltage is zero, therefore I'm trying to understand how the formula on the picture has been derived.

2 diodes in serires with a resistor

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  • \$\begingroup\$ What are the circles between the diodes and \$R_D\$'s? \$\endgroup\$ – Null Apr 20 '15 at 18:40
  • \$\begingroup\$ Why do you say that it's missing? It's right there on the bottom. \$\endgroup\$ – Greg d'Eon Apr 20 '15 at 18:41
  • \$\begingroup\$ They represent the voltage drops across the each diode. Why they are positive baffles me. Sry see the U and R for for R, edited initial question. \$\endgroup\$ – BoroBorooooooooooooooooooooooo Apr 20 '15 at 18:41
  • \$\begingroup\$ (removed comment on what is a convention) When you compute the current in a loop you account for all generators in the loop (Um and the two Ut's) and not for voltage drops on resistors. The resistors are accounted for in the denominator. \$\endgroup\$ – Sredni Vashtar Apr 20 '15 at 18:43
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If you rearrange your formula, you get $$ (U_m - 2U_T) = I_m (2R_d + R) $$ which looks like $$ V = IR $$ where \$R\$ is the sum of all of the resistors in the loop and \$V\$ is the voltage across that summed resistor.

Kirchoff tells you that the voltage across this resistor is the same as the voltage looking at the rest of the circuit, which is \$U_m - 2U_T\$, because there must be a voltage drop across each of the diodes for current to flow.

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