circuit with four resistors and voltage source

I need to completely understand how to find the value of each resistor in this circuit. I understand that with R1: i1 and V1 flow in the same direction and Ohm's Law can be directly applied but could someone help me with R2. Current is flowing from negative to positive which is opposite to the voltage. Do I perform a sign change to the i2 which for this exercise is the same as i1 (-e^(-2t)) A?

  • \$\begingroup\$ The reversed voltage on R2 just means that someone's guess about the polarity was wrong. \$\endgroup\$ Commented Feb 26 at 0:37
  • 3
    \$\begingroup\$ @PeterBennett but placing the i2 arrow to go into the negative terminal of the resistor means they're ignoring the passive polarity convention (and the other devices don't consistently use the active polarity convention). OP: My advice is re-draw the circuit and solve it with your own choice of symbols, then convert them to the ones shown in the diagram when you present your answer. \$\endgroup\$
    – The Photon
    Commented Feb 26 at 1:10

2 Answers 2


You can think of the polarity signs on the resistor as the problem telling you how to attach a multimeter across the resistor to measure it, with the red lead on the + symbol and the black lead on the - symbol. It doesn't "set" the voltage developed across the resistor, it just says that the quantity V2 is defined as the voltage at the + symbol minus the voltage at the - symbol.

Your intuition tells you that current is flowing out of the + terminal of the voltage source and back into its - terminal (which is correct in this example; get ready to see currents flowing into the + terminal of voltage sources) and therefore the "assigned" polarity on R2 doesn't match the voltage you know will be developed across its terminals; V2 will be a negative quantity. This is fine, all the math you care to do with these quantities will still "work" correctly. When you start doing more complex analysis you will not be able to eyeball the circuit and assign directions and signs to all the currents and voltages so learning how to provisionally assign them and roll with positive and negative signs is a key skill and part of what this problem is trying to get you comfortable with.


This is so confusing, but I'm sure that is the purpose of this exercise. When passive sign convention is ignored, as it is in the cases of R2 and R4, the voltage across the element will have the wrong sign.

So yes, you are correct when you suggest that V2 should appear negated in your equations. You will have the same problem with V4.

If you apply KVL to the left loop, starting at bottom-left, going clockwise in the labelled current direction, and you stick to voltage polarities as they are labelled in that diagram, this is the equation that emerges:

$$ v - v_1 + v_2 - v_3 = 0 $$

This will yield the wrong sign for \$v_2\$, because conventional current always enters a resistor at the end with the higher potential, and exits the resistor from the lower potential terminal. That is reality, and is what passive sign convention aims to reflect.

For the simultaneous equations to yield the correct magntude, that KVL equation should negate \$v_2\$:

$$ v - v_1 - v_2 - v_3 = 0 $$

However, when you finally obtain a value for \$v_2\$, it will have the correct magnitude but the wrong sign, in the context of this diagram, and you'd have to keep this in mind as you proceed.

It's better to redraw the diagram, in accordance with passive sign convention, before proceeding. I've made this correction for both \$v_2\$ and \$i_4\$ here:


simulate this circuit – Schematic created using CircuitLab

Then you can derive the equations and solve them, with confidence that it will yield the correct signs and magnitudes, for the new diagram. However, you must remember to negate your results for \$v_2\$ and \$i_4\$ for them to be correct in the context of the original schematic.


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