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.
Best Answer
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.