# Is my equivalent circuit, with respect to the open circuit, correct?

Target: Finding an equivalent circuit with respect to the open circuit, not the closed one. My 1st question is whether or not my equivalent circuit is correct or not. I'm confused whether or not I should add up resistors even if the circuit is opened.

My 2nd question is: Why when I apply KVL on left loop I end up with a value of V different from the value I get using KVL on right loop. Is my mistake that I have to find equivalent circuit first then apply KVL to find V? Why?

Thank you

• "Equivalent" has meaning only when it is defined where one looks at it from. Commented Jan 17, 2018 at 22:07
• I don't believe so since the "original" circuit has a closed path and your "equivalent" does not. Commented Jan 17, 2018 at 22:26
• @RonBeyer Norton equivalent of Voltage source with series resistance is closed. Commented Jan 17, 2018 at 22:28
• No, I want to find an equivalent circuit with respect to the open circuit.
– Joe
Commented Jan 17, 2018 at 22:29
• @Eugene Sh.: obviously the two terminals across the switch are meant. I.e. SC current (current through switch when closed) and OC voltage (voltage across switch when open) are equal iff both circuits are equivalent.
– Curd
Commented Jan 17, 2018 at 22:29

Verify it yourself:

The circuits are equivalent (with respect to the two terminals of the switch) iff

• short circuit current (through switch when closed) and
• open circuit voltage (across switch when open)

are the same.

• What about my 2nd question
– Joe
Commented Jan 17, 2018 at 22:35
• @Joe: You asked "when I apply KVL on left loop I end up with a value of V..." but how do we know what calculation you did and what value for V you ended up with??? At least I don't have the gift of clairvoyance.
– Curd
Commented Jan 18, 2018 at 8:06

No, your circuit is not right because your task was to find the equivalent circuit with respect to the open switch. So that basically means you could ignore the part of your circuit where the open switch is, because there will be no current, which means that this part is irrelevant.