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WesT4N

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## Homework Statement

Given a battery with voltage 35.7 V and 4 Resistors (Top left with resistance 2R, and the 3 others each with resistance R) where R = 507 Ohms, what current would an ideal ammeter read in its current position? (Diagram attached)

## Homework Equations

Kirchoff's Laws: KCL and KVL

Ohm's Law: V = IR

Resistors in series: R_eq=R1 + R2 + ...

Resistors in parallel: 1/Req = 1/R1 + 1/R2 +...

## The Attempt at a Solution

I first attempted to assign current variables (I1 leaving from battery, splits into I2 and I3 at the junction, and then current is redistributed again into I4 (bottom left) and I5, with I6 being the current at the ammeter's location).

Since the resistor at the top right has less resistance than one at the top left, more current should flow through the right side, thereby making I6 flow from right to left.

I then apply Kirchhoff's Laws and get a system of equations.

I2 + I6 = I4

35.7 - I2 * 2(507) - I4 * 507 =0 (Running from Battery, looping clockwise with I2 and I4)

35.7 - I3 * (507) - I4 * 507 = 0 (Running from battery, looping clockwise with I3 and I4).

-I2 * 2 (507) + I3 * 507 = 0 (Looping in the top right section).

No matter what combination of loops I try, I seem to always have one free variable, which indicates that I'm missing a relationship between the currents.

Does I3 = 2 * I2?

Am I correct in viewing this as a pair of parallel resistors in series where I'm trying to find the current coming out of the first pair and into the second?

Any help is appreciated.