David K. answered • 03/25/22
Expert, Friendly Physics Tutor with 5000+ Hours Tutoring Experience
Hi Josh -
Without a complete diagram of the circuit, it's a little difficult to be certain what the answer to this problem will be. Assuming that our circuit is just a simple loop of wire beginning at the battery, passing through the three resistors followed by the ammeter in series (i.e. one after the other without anything branching off of the main loop in the circuit), and then reconnecting to the battery to complete the loop, we can figure this one out.
The big idea that you need to remember to solve problems like this one that involve resistors in series is how to calculate the equivalent resistance (that is, the resistance of a single resistor that could be used to replace the three resistors in this problem without changing the total resistance in the circuit). For resistors in series, we calculate equivalent resistance by adding up the individual resistances. In this problem, the equivalent resistance is:
3Ω + 6Ω + 9Ω = 18Ω
meaning that we could remove the 3 resistors we are given in this problem, replace them with a single 18Ω resistor, and the circuit would behave the same. Now that we know the equivalent resistance and the voltage of the battery, we can apply Ohm's Law to find the current, which says
V = IR
where V is the voltage of the battery powering the circuit, R is the equivalent resistance in the circuit, and I is the current that will flow through the circuit (the thing we are trying to solve for). Plugging in our known values, we get:
36 volts = I * 18 ohms
2 amps = I
Dividing both sides by 18 yields our answer: the current in this circuit will be 2 amps (note how the units work out: dividing volts by ohms gives us a quantity with units of volts per ohm, which is the same as amps).
I hope this helps - let me know if you have any other questions and I'll be happy to explain more!
