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**Circuit**: Here's the circuit: there is a 8V battery. The charge leaving it reaches a junction: if the current "turns right" it goes to a 12 ohm resistor (R1) and can then loop around back to the battery, if it goes straight (i.e. takes other path at junction) it runs into a 4 ohm resistor (R2). After the 4 ohm resistor, there are two more 4 ohm resistors (R3 and R4) in parallel (i.e. the equivalent resistor of these two is in series with the first 4 ohm resistor).

**Question**: Which resistor has the most current going through it?

**Relevant equations**: resistors in series and in parallel, Kirchoff's loop law

**Attempt at solution**: If I solve the equivalent resistor for the two 4 ohm resistors in parallel, it's 2 ohms, which is in series with the other 4 ohm resistor, giving 6 ohms. This is less than 12 ohms, so the path "straight ahead" described above (i.e. NOT the path leading to the 12 ohm resistor) is the path of least resistence. Since the first 4 ohm resistor is in series with the equivalent resistor of the other two that are in parallel, then they must have the same current flowing through them by the loop law.....

So I was thinking that R2, R3 and R4 would all have the same current through them, but my multiple choice options only include R2, R3, and R4 individually or R3 and R4 together.

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