We just started learning about diodes and now I want to find the currents \$I_1\$, \$I_2\$ and \$I_3\$ in this circuitenter image description here. We should assume that the amount of voltage the diodes recieve is 0,7V each and that they got no resistance.

I worked this through:

For I1 I got: \begin{equation}I_1 =\frac{(U-2\cdot 0,7V)}{\frac{R_1\cdot R_2}{R_1 + R_2}}= 5,003\,mA\end{equation}

For I_3: \begin{equation}I_3 = \frac{U-2\cdot 0,7V}{R_1} = 2,268\,mA\end{equation}

And for I_2: \begin{equation}\frac{U-2\cdot 0,7V}{R_2} = 2,735\,mA\end{equation}

However, I am not sure if those values are correct.

  • \$\begingroup\$ Hi @Eren for inline Latex markup you need to put \$ \$\endgroup\$
    – crowie
    Dec 3, 2016 at 12:39
  • \$\begingroup\$ You are on the right track with you calculations but you have made a couple of wrong assumptions. R1 and R2 are actually in series not parallel. Hint: Work out what I2 is first then I3 then add both together to get I1. R1 is a part of both I2 and I3 and hence I1 \$\endgroup\$
    – crowie
    Dec 3, 2016 at 12:46
  • \$\begingroup\$ Thank you very much. So since \$R_2\$ is parallel to the second diode, this means that the voltage at \$R_2\$ must be 0,7V right? With that, I can calculate \$I_2\$ , and \$I_3\$ I can compute by the same way as before? \$\endgroup\$
    – Eren
    Dec 3, 2016 at 12:54
  • 1
    \$\begingroup\$ Exactly you got it \$\endgroup\$
    – crowie
    Dec 3, 2016 at 12:57

1 Answer 1


I won't answer this fully but you have to treat each path separately to get the correct answer. But heres 1 path you should be able to work out the rest yourself. We work on the assumption the diode forward voltage drop is 0.7V. $$ I_2 = 0.7/6800 = 0.103mA $$


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