Consider the circuit shown in the figure below. (Assume R1 = 12.0 Ω and R2 = 6.00 Ω.)

Read this for the figure -

A rectangular circuit begins at its left side at a 25.0 V battery with the positive terminal on top. From the battery, the circuit extends up and to the right to point b. The circuit continues rightward from point b and splits into three parallel vertical branches. The leftmost branch has a resistor labeled R1, the middle branch has a resistor labeled R2 and the rightmost branch, from top to bottom, goes through a resistor labeled R2, point c, then and a resistor labeled 20.0 Ω. The three branches recombine, the circuit extends to the left passing through point a, bends upward to a resistor labeled R1 and reaches the negative terminal of the battery.

(a) Find the potential difference between points a and b.
(b) Find the current in the 20.0-Ω resistor.

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  1. Over on the right you have three branches
    in ohms
    left 12
    middle 6
    right 6 + 20 = 26
    find Re or equivalent resistance of those three right legs
    1/Re = 1/12 + 1/6 + 1/26 = .08333+ .16666 + .03846 = 0.28845
    so Re = 3.467 ohms
    resistance around the circuit = 3.467 + 12 = 15.47 ohms
    so total current = 25/15.47 = 1.61 amps
    so voltage drop across bottom R1 = 1.61 * 12 = 19.4 volts
    so voltage drop across Re = 25 -19.4 = 5.61 volts (that is answer 1)
    so current through rightmost leg = 5.61 / 26 = .0235 amps

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