Some Practical Consequences

Consider the following question.
Two light bulbs (which we'll treat as Ohmic resistors) are connected in parallel with a battery, as shown. The brightness of each light bulb is related to its power output, which is proportional to the current through the bulb ($$P=I^2R$$). If I unscrew the 2Ω light bulb (that is, if I remove the bulb from the circuit, leaving a gap where it was), the 4Ω bulb
gets brighter stays the same
gets dimmer goes out
Some folks would look at this problem and think If I unscrew the 2Ω light bulb, then all of the battery's current will go into the 4Ω bulb instead, and so it must be brighter.

The loop rule, however, tells a different tale. Tracing a loop through the battery and the 4Ω bulb gives us the equation $$8-4I_2=0 \implies I_2=2\u{A}$$ This answer doesn't depend on the 2Ω bulb at all!

It is true that, when you unscrew the 2Ω bulb, the current $$I_1$$ no longer flows through it. But that current isn't diverted into the other bulb; instead that current simply stops flowing. Remember that the battery doesn't provide a constant current or a constant energy; it maintains a constant voltage, providing as much or as little as current to do so. In the extreme case, if I unscrew both light bulbs, then the battery provides no current at all.

In most circuits, every component affects every other. The circuit in the example above is unusual in that the light bulbs are independent of each other. In fact, if I added a third light bulb in parallel to the first two, or a fourth, this independence would remain.

This independence is highly desirable when you wire a house. When I turn on a light, I don't want the other lights in my house to dim. Turning off the television shouldn't send a surge of power into my laptop. Houses are wired in parallel for this reason. (Or as closely as possible… we'll talk about that in the next example.)
Suppose the 8V battery in the previous example has an internal resistance of 1Ω. If I unscrew the 2Ω bulb, the 4Ω bulb
gets brighter stays the same
gets dimmer goes out
Now wait, I thought resistors in parallel are independent! No, that's only if the resistors are also in parallel with a battery. Using the same loop from the previous example gives us $$8-4I_2-1I=0 \implies I_2=2-\frc4I$$

The current $$I_2$$through the second bulb now depends on the current $$I$$ out of the battery, and the current out of the battery will depend on what else is connected to the battery. When I remove the first light bulb, the battery needs to provide less current, so $$I$$ gets smaller, and $$I_2$$ gets bigger.

It may be more convincing to solve for the currents. With both light bulbs in place, the following equations are true: $$-2I_1+4I_2=0 \implies I_1=2I_2$$ $$I=I_1+I_2=2I_2+I_2=3I_2$$ $$8-4I_2-1I=0 \implies 8-4I_2-2I_2=0 \implies \boxed{I_2=1.33\u{A}}$$ $$\implies I_1=2I_2\implies \boxed{I_1=2.67\u{A}}$$ $$\implies I=I_1+I+2\implies \boxed{I=4\u{A}}$$ When the 2Ω bulb is removed, then $$I_1=0$$ and $$I=I_2$$. Now there's only one unknown, $$I$$, and a single equation will suffice: $$0=8-4I_2-1I=8-4I-1I \implies \boxed{I=1.6\u{A}} \hbox{ and } \boxed{I_2=1.6\u{A}}$$ The current through the 4Ω resistor goes from 1.33A to 1.6A, and so that bulb gets brighter.

The larger the internal resistance of the battery, the greater this effect is; and the same is true in house wiring. The wires that bring electricity into a house provide internal resistance, and other factors contribute as well. If the light bulbs (or televisions, or appliances) in a home all have resistances which are much larger than the internal resistance of the power supply, then the internal resistance doesn't matter much, and the devices will act independently, as in Example 11.6.1. Add a device with low resistance, however, and the internal resistance becomes significant. Recall   TBD   that a device plugged into a $$110\u{V}$$ outlet uses power equal to $$P=(110\u{V})^2\over R$$; so a "low-resistance device" is a high-power device, like a vacuum cleaner or a microwave. Turn one of these on, especially in an older house whose power supply has high internal resistance, and the devices are no longer independent. Instead, we see the result form Example 11.6.2: the lights dim.