0
\$\begingroup\$

In the schematic below I was wondering why my professor added Vs(0-)/s when doing KVL. When I applied KVL I had Vs(s)=RI(s)+I(s)(1/cs)-Vc(0-)/s. The reason I subtracted was because after assuming current flows clockwise that meant current would enter Vc(0-)/s through the positive end and go out through the negative end, meaning I'd have to subtract. Why did he add and not subtract?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
  • \$\begingroup\$ if you put the voltage sources on the left side it becomes negative due to orientation. Or you sum all equal to zero then Vs becomes negative \$\endgroup\$ Jul 11, 2020 at 22:46
  • \$\begingroup\$ That's a weird notation. What's Vc(0-)/S supposed to mean ? \$\endgroup\$
    – Hilmar
    Jul 12, 2020 at 16:26

2 Answers 2

1
\$\begingroup\$

Ignore the resistor altogether and write an equation with both \$V_c\$ and \$V_s\$ terms on the same side of the equal sign. Should they be the same sign? Or opposite?

Put another way, take the equation in your opening post and move Vc to the left-hand sign of the equation. That gives you \$V_s + V_c = [...]\$. Does that make sense from looking at your circuit? It should not make sense since the equation says the two voltage sources add up but in the circuit you can clearly see they are facing opposite directions.

You do not choose positive or negative based on whether the component is a source or load. The reason is because you can have things like voltage sources facing opposite directions like in this circuit, and the direction you assume the current flows around the circuit is also arbitrary (sometimes you don't know if something is a load or a source).

You choose positive or negative based on whether the voltage falls or rises when tracing the current through the component. Look at your resistor, is the voltage rising or falling as the current goes through the resistor? In other words is the current entering the + or - voltage side of the resistor? Now compare that polarity with the voltage drop polarity across \$V_c\$. Is it the same as the resistor? Or different? If it it's the same, the sign for the \$V_c\$ term should be the same as the \$R_1\$. If it's different it should be the opposite sign.

Basically, if all terms are positive in your equation, everything on one equal sign is opposing everything on the other side of the equal sign. If there is a negative term in there, it means that term is opposing everything on the side it is on. When starting out, this is much more obvious if you just go \$0 = \sum{V}\$

I recommend you do \$0 = \sum{V}\$ for your equation when starting out so you get used to when a term should be negative and positive. Trying to do \$V_{source} = \sum{V_{load}}\$ gets you thinking too visually and in the bad habit of trying to guess and assign what you think is the "true" source of the circuit. But this can trick you since sources can act as loads for other sources if they face each other in opposite direction. This can get really bad in circuits with multiple loops where you have no idea.

For example, in your original equation what is so much more special about Vs than \$V_c\$ that made you choose to put \$V_s\$ all by itself on one side of the equation? There is no reason, yet that's what you did and it tripped you up.

\$\endgroup\$
0
\$\begingroup\$

You are wrong. Think the case Vs is a constant DC and Vc is initially the same as Vs, so there's no more room for any current different from zero.

Transform your equation to form I=something. Your minus gives plenty of room for the current. Professor's version shows that I=0 which is right.

\$\endgroup\$

Your Answer

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

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