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

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.

Suppose the 8V battery in the previous example has an internal
resistance of 1Ω. If I unscrew the 2Ω bulb, the 4Ω bulb
*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
$$8-4I_2-1I=0 \implies I_2=2-\frc4I$$

gets brighter
stays the same

gets dimmer goes out

gets dimmer goes out

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.