A 9 V battery has its negative terminal connected to ground. Two separate paths run from its positive terminal back to ground: a right-hand path made of a 20 ohm resistor and a 100 ohm resistor wired one after the other, and a left-hand path made of a single 50 ohm resistor. How much current flows in the right-hand path?

1. The right-hand path's two resistors are wired one after the other, so their resistances add: R_right = 20 ohm + 100 ohm = 120 ohm. 2. Both paths run between the same two points, the battery's positive terminal and ground, so both paths see the full 9 V of the battery. 3. Apply Ohm's law to the right-hand path alone: I_right = V / R_right = 9 V / 120 ohm. 4. This is a series-inside-a-branch situation: the two resistors combine by simple addition even though the two paths themselves are separate branches. 5. Using 100 ohm alone ignores the 20 ohm resistor also present in that same path. 6. Using 50 ohm applies the other branch's resistance to this branch, which is not connected to it. 7. Using 20 ohm alone ignores the 100 ohm resistor also present in this path. _Source: OpenStax Physics (CC BY 4.0), Ch 19 "Electrical Circuits", section 19.2 Series Circuits_