# Determine I through load resistor in circuit with diodes in series

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. • What are the circles between the diodes and $R_D$'s? – Null Apr 20 '15 at 18:40
• Why do you say that it's missing? It's right there on the bottom. – Greg d'Eon Apr 20 '15 at 18:41
• 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. – BoroBorooooooooooooooooooooooo Apr 20 '15 at 18:41
• (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. – Sredni Vashtar Apr 20 '15 at 18:43

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.