Question

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.

Answer 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