How do I solve this simple circuit using Kirchoff's Laws?

Question

I am a freshman taking up Electronics Engineering and I am brushing up my skills in doing some simple circuit analysis problems. However, I've come across this quite simple problem and my answer is different. The book I am reading says that to do KVL, the polarity of the terminal encountered first must be the sign of the voltage. I followed that and my answer is apparently wrong. May I ask where did I go wrong?

Answer 1

Using KVL sounds easy, but polarities can really mess you up.

The approach I usually advise is that one does not take into consideration polarities when traversing all the elements in a loop, and that we rely on the signs of the resulting solution to reveal actual polarity. This entails adding each term, regardless of polarity, and then afterwards, when plugging in all the known values, we set the sign of those knowns to be consistent with the direction we travelled.

Alternatively, we can take the approach where if we do know the polarity of voltage across some component, we account for drops in voltage as we travel over it by subtracting, and we account for rises in potential by adding. I'll take that approach here, so you can appreciate what I mean.

I'll redraw the circuit here, with labels so the algebra makes sense:

^{simulate this circuit – Schematic created using CircuitLab}

Adding and subtracting while traversing the loop

Here I apply KVL by adding when I see a rise in potential, and subtracting when I see a fall in potential as I traverse each element. My thinking goes like this:

Starting at the bottom left, where I arbitrarily "declare" the potential to be zero volts, going clockwise, I first cross \$V_a\$, which entails a rise in potential, so I add. My next leg of the journey is across R1, where I detect a drop in potential, as the authors have kindly told us will be the case, with those signs, so I will subtract \$V_1\$. Then I jump over \$V_b\$, encountering another rise in potential as I go, so I add again. Lastly I traverse R2, and the signs tell me that potential will rise, so I add \$V_2\$. I'm back to where I started now, which must be at the same potential (zero difference) as it was when I began the journey. I can set the entire expression to 0V. I have constructed this equation:

$$ 0V + V_a - V_1 + V_b + V_2 = 0V $$

Now we apply KCL, with the simple observation that the current must be the same all around the loop. We are ready to apply Ohm's law to the resistors:

$$ V_1 = i \times R1 = 4i $$

It's absolutely crucial at this point to notice that as we applied Ohm's law to R1, we didn't break the condition that current must flow from the end with the higher potential to the lower potential. The "+" and "–" signs are consistent with that requirement.

However, for R2, this is not the case. You can see that current is flowing from lower to higher potential, if those "+" and "–" signs are to be believed. That means obviously the authors have got this wrong. That's OK, but we must apply Ohm's law in a manner consistent with reality, and either the voltage or current must be "reversed" or negated here:

$$ V_2 = -i \times R2 = -2i $$

Plugging these expressions for \$V_1\$ and \$V_2\$ into the KVL equation gives us:

$$ V_a - 4i + V_b + (-2i) = 0V $$

Plugging in our known values for \$V_a\$ and \$V_b\$, and solving for \$i\$:

$$ 10 - 4i + 8 - 2i = 0 $$ $$ -4i - 2i = -8 - 10 $$ $$ -6i = -18 $$ $$ i = 3A $$

The final solutions for \$V_1\$ and \$V_2\$ are:

$$ V_1 = 4i = +12V $$ $$ V_2 = -2i = -6V $$

The negative value for \$V_2\$ is simply telling us that our assumption about the polarity across R2 was wrong, but I want you to notice that the arithmetic still worked out. The only potential for error here (pun intended) was that we may have missed the inconsistent application of Ohm's law for R2. Be very careful in this respect. It might be prudent at this point to reverse the "+" and "–" symbols in the schematic.

Answer 2

When in doubt, simulate it out...

^{simulate this circuit – Schematic created using CircuitLab}

So, the answers are correct.

Now, how are they derived?

Well, we can rearrange thing to make it even simpler.

^{simulate this circuit}

Now, it should be pretty obvious that 3 A is flowing in the circuit. So, voltage across R1 is 12 V and R2 is 6 V, however, they put the measurement polarity signs on R2 opposite, so it is -6 V.