# Applying kirchoffs law in a simple circuit [closed]

Can we apply Kirchhoff's volatge law in this simple circuit and say V1=-r1I1 and V2=r2I2 or is there an other connection between voltages and currents?

• It all depends on what those circles represent. Commented Jun 19, 2020 at 13:27
• @Andyaka Thanks for the answer sir!They are voltage sources. Commented Jun 19, 2020 at 13:30
• @user287001 Woah dude you are scaring me please stop it Commented Jun 19, 2020 at 21:50
• I can well remove the comment but that doesn't create any solutions. Removed.
– user136077
Commented Jun 19, 2020 at 21:52
• Homework needs an attempt at a solution Commented Jun 26, 2020 at 19:23

Kirchhoff's Laws are not needed here. If you have an ideal voltage source, and the source value is given (whether a literal value or a function of other circuit values) then you can state the voltage across that source by inspection.

• So could we say V1=-I1r1? Commented Jun 19, 2020 at 13:39
• The polarity of V1 is defined with a direction that is opposite of the defined polarity of the source on the left. Commented Jun 19, 2020 at 13:47
• How could you relate V1 to V2 then? Commented Jun 19, 2020 at 14:54
• There is no relationship between the values of V1 and V2 as they are drawn in this diagram. They are independent voltages. Commented Jun 19, 2020 at 15:12

Yes. Kirchoff’s voltage law is a law and can always be applied around a loop in a circuit. If you’ve defined your polarities correctly KVL will give you a correct answer

• And is the correct answer the one i suggested? Commented Jun 19, 2020 at 13:32

Judging from their drawing, OP is studying two-port networks.

That said, OP's task may be to find impendence, admittance, or hybrid parameters, and r_1, r_2 are not given values, but the values to be calculated. If so, and if the sources are independent, Ohm's law is sufficient for solving this simple problem: $$z_{11} = {-V_1 \over I_1}\\ z_{22} = {V_2 \over I_2}\\ z_{12} = z_{21} = 0\\ (r_1=z_{11}; r_2=z_{22})$$