# How to write KCL and KVL in these circuits?

I want to use KCL on this circuit. I have $i-i^{i}_{D2}-i^{i}_{D1}=0$.

Are the correct currents:

$i=10V/R_1$?

$i^{i}_{D2}=V^{i}_{D2}/R_2$?

But that is $i^{i}_{D1}$? I have no resistor.

On the circuit below I want to use KVL, but how should I do that? I don't know how to include $V^{i}_{D1}$ in KVL.

I guess this is wrong $10V-R_1i-V^{i}_{D2}-R_2i^{i}_{D2}-(-10V)+10V=0$?

Thanks!

There's a voltage difference across the resistor R2.

You can Omit $V^{i}_{D2}$because $V^{i}_{D2}$= 0.

So, voltage across $R_{2}$ is = 0V - ( -10V) = 10V.

So, $i^{i}_{D2}$= 10V / $R_{2}$ = 10V / 5000 Ohms = 2 mA.

As you said: $i$ = 10V / $R_{1}$ = 1 mA.

You can substitute the values of $i^{i}_{D2}$ and $i$ in the equation of KCL to get $i^{i}_{D1}$. That's because you don't have a resistor. So, I think this is the easiest way to calculate $i^{i}_{D1}$.

$i^{i}_{D1}$= $i$ - $i^{i}_{D2}$= -1 mA. (Negative means opposite direction).

You said $i^{i}_{D2}$= $V^{i}_{D2}$ / $R_{2}$ = 0 !! That's not true. Because $V^{i}_{D2}$ is not the voltage across R2. It is the voltage across the diode only (or the switch).

For the second question:

Edit: You should choose a smaller loop to include $V^{i}_{D1}$ such as:

$-10V - V^{i}_{D1} + V^{i}_{D2} + i^{i}_{D2} * R_{2} = 0$

Since there's no voltage across points A and B as you demonstrated in the picture, This means: $V^{i}_{D2} = 0$ and also $V^{i}_{D1} = 0$.

This makes the equation simpler: $-10V + i^{i}_{D2}$ * $R_{2} = 0$

or $i^{i}_{D2} = 10V / R_{2} = 0$

For the outer loop: Think of 10V and -10V as a battery of 20V and this battery feeds the two resistors so,

20V = $i$ * $R_{1}$+ $i^{i}_{D1}$ *$R_{2}$

• Thank you! What are the equations for the voltage across $R_2, V_{D1}, V_{D2}$? (The last two are zero, but I wondering out of curiosity). I don't have any potentials at point A and B (Marked in this picture: ibb.co/gBixJv) Commented Apr 2, 2017 at 12:28
• Might be worth modelling on everycircuit.com (its a free tool and its really visual) Commented Apr 2, 2017 at 15:13
• @JDoeDoe Oh I'm sorry there was a mistake. I've edited my answer. I edited the equation of the smaller loop to include R2, VD1 and VD2. Commented Apr 2, 2017 at 18:02

Are the correct currents:

$$\ i = 10V / R_1 ?\$$

yes.

$$\i^{i}_{D2}=V^{i}_{D2}/R_2 ?\$$

No. $$\i^{i}_{D2}= 10V / R_2.\$$

As $$\i = i^{i}_{D1}+ i^{i}_{D2}.\$$ solving for $$\i^{i}_{D1}\$$ is easy.