Using Ohm's Law and simplification of the circuit, what are the resistor, power, current, and voltage values?

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A simplified circuit would eliminate R4 by adding its value to R3, and putting that in parallel with R2. Everything to the right of point B is eliminated, no function in a circuit.
You could then simplify it further by working out the equivalency of the 2 paralleled resistances, leaving that in series with R1 across the battery.
In the end, you could add the equivalent value to R1, and represent the circuit with just 1 resistor in series with the battery.
R4 and R3 in series = 200 ohms. This is in parallel with R2 of 200 ohms, so equivalent resistance is 100 ohms.
a) This is in series with R1 of 100 ohms, to final circuit resistance = 200 ohms, and the fully simplified circuit would be a 200 ohm resistor across the 3V. battery.
b) Current I = (E/R) = (3/200) = 0.015A.
c) Power = (EI) = (3 x 0.015) = 0.045W.
d) R1 would pass all the circuit current in the unsimplified circuit, = 0.015A.
e) Voltage relative to ground = 3V. at E.
f) V at A = 3v - (V drop acrossR1) = 3 - (0.015A x 100 ohm) = 1.5V.
g) The 1.5V. is between C and A, so current through r2, I = (E/R) = (1.5/200) = 0.0075A.
h) Power dissipated in R2 = (EI) = (1.5 x 0.0075) = 0.01125 W.
i) R4 and R3 have a current of (0.015 - 0.0075) = 0.0075 A. passing through both.
j) Voltage at B = 1.5 - (IR) = 1.5 - (0.0075 A x 60) = 1.05V.
k) It will be 1.05V. relative to ground, but all that circuit after point B is not doing anything.
l) Source 0.1A., voltage at E would be (IR) = (0.1 x 200) = 20 V. (200 ohm is the equivalent resistance of the circuit)....Show more

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