1
\$\begingroup\$

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.

\$\endgroup\$
4
  • \$\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

Reset to default
0
\$\begingroup\$

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. $$ I_2 = 0.7/6800 = 0.103mA $$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.