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}\$
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.
