Kirchhoff’s Loop Rule doesn’t apply to an entire circuit?

Question

True or false: the sum of the voltage sources in a circuit is equal to the sum of the voltage drops in that circuit

That is the statement which was presented. I thought it was strange because it was almost verbatim from the explanation in the section I was reading. The only difference was that the reading stated “around any closed loop, the sum of...” and this statement said “in a circuit” instead of “a closed loop”. As far as I understand, a circuit is a closed path in which current/electricity flows, and Kirchhoff’s loop rule is just a specific scenario where energy conservation is applied. Therefore I’m confused why this is false because if a circuit is a closed path, wouldn’t Kirchhoff’s law apply to it?

Also I’m not sure if this will help but this is from the answer key:

While the voltage sources and voltage drops are equal in any closed loop, this is not necessarily true for the entire circuit. For example, a 9V battery that powers 10 light bulbs in parallel has a 9V voltage source and a 9V drop across each light bulb - a total of 90V of drop across all of the light bulbs combined.

I could only comprehend the first sentence but I don’t understand why that is the case. Also, if a total of 90 V of across all the light bulbs occurs, wouldn’t there also have to be 90 V of voltage source? If there was more source, energy would’ve randomly appeared and if there was less, unaccounted-for energy would’ve dissipated. In either case, wouldn’t the law of energy conservation be violated?

Edit: I’m not sure why the answer key gives the example with the parallel circuits because the book didn’t cover them in the section this statement was in. I haven’t looked at parallel and series resistors in a while so I’d appreciate an explanation focused more on the loop rule.

Answer 1

True or false: the sum of the voltage sources in a circuit is equal to the sum of the voltage drops in that circuit

The statement is misleading. Per KVL the sum of the voltage rises in a circuit is equal to the sum of the voltage drops in the circuit. But a voltage rise is not necessarily due to a voltage source, as in a battery or other source of electrical energy.

Refer to the circuit diagram below.

Applying KVL to loop 1 proceeding clockwise we obtain:

$$+42-12I_{1}-6I_{1}+6I_{2}=0$$

Notice that current $I_{2}$ contributes a voltage rise of $6I_{2}$ in loop 1, yet it is not a voltage source in the sense it is not an active voltage source, but merely a voltage drop in loop 2. On the other hand the voltage rise of 42V in loop 1 is due to a voltage source, the battery.

Hope this helps.

Answer 2

The problem is that, when things are connected in parallel, the "sum of the voltage drops" across them doesn't really mean much. Each of the lightbulbs will have a 9V voltage drop across it, so if you naively "sum" them up, you get a number equal to 90V. The point is that this number doesn't have much physical significance.