Originally Posted by **rcmango** from the diagram i drew I want to calculate the current in the battery for the circuit, can someone help me with this. Want to use symbols too. R1 and R2 are in series, they are in parallel with R3, and R4 is in series with R1, R2, and R3. The diagram is attached. |

This is a problem, like any other involving resistors and a DC battery that is best solved by always combining the elements into as few as possible. So you have R1 in series with R2, both in parallel with R3. Let's add R1 with R2 getting 10 Ohms, and then put that in parallel with R2 which is:

Rc = Resistor combination = 1/[1/(R1 + R2) + 1/R3]

Rc = 1/[(1/10) + 1/(15)]

Rc = 1/(0.1 + 0.06666) = 6 Ohms

Now you have a circuit that is equivalent to the one you had with only two resistors in it. They are both in series and can be summed as such.

One has a value of Rc = 6 Ohms and the other has a value of R4 = 8 Ohms.

The total resistance the battery sees to sourcing current into the circuit is then: Rc + R4 = 8 + 6 = 14 Ohms.

You do not actually have a battery voltage written in your picture so I will just call it V.

The current that is sourced from the battery is:

I = V/14 amps

Whenever you have a circuit that has multiple combinations of resistors in parallel and in series, and you wish to find the total current being sourced from the battery, it is always safest to simply combine the resistors into a single equivalent resistor the battery sees. Mathematically combining them into one and replacing all the others with that one is no different to the battery than using all the series parallel combination resistors in the original circuit. It still "sees/feels" the same resistance and so it will source the same current. It doesn't know the difference between a single resistor of an equivalent resistance compared with many different resistors who series and parallel resistance combination is the same as that single one.

Take care,

Craig