Physics! I can calculate the electric current flowing through a complex circuit based on resistance and voltage.

Question

solve this and fill in the table below please provide all the steps.

Especially on how you found voltage, Current . Thank you

(Total: 12V, 3A, 4Ω)	Voltage	Current	Resistance
Total	12V
R1			4Ω
R2			4Ω
R3			2Ω

Answer 1

Ohm's law: V = I*R

Series:

1. R_s = ∑R_i
2. I is constant
3. V gets divided over resistors so that V_i = I*R_i, with V = ∑V_i

Parallel:

1. R_p = (∑(1/R_i))^-1
2. V is constant
3. I gets divided over the resistors/branches such that V = I_i*R_i and I = ∑I_i

We've got a combination of parallel and series here. The values for the entire circuit are already given in your table but let's go over the calculations.

R₁ and R₂ are in parallel and that parallel part is connected in serie with R₃.

We first substitute the parallel part with a resistance R_p = (1/R₁ + 1/R₂)^-1 = 2 Ω

The total resistance or equivalent resistance of the circuit is then R = R_p + R₃ = 4 Ω (series).

The total current in the circuit is I = V/R = 3 A.

Let's retrace on our steps.

We have a series circuit with R_p and R₃ with a current I and voltage V.

In series, the current is constant (it can only follow one path), I_p = I₃ = I = 3 A

The voltage gets divided over both resistors R_p and R₃, who happen to have the same value,

so the voltage gets divided evenly: V_p = V₃ = 6 V. If this was not the case you can always calculate

V_i = I*R_i, (in this case both will be 3 A * 2 Ω = 6 V).

Another step back, let's look at the parallel part of the circuit.

We now know that the voltage over this part is V_p = 6 V and the current is I_p = 3 A.

In a pure parallel circuit the voltage is constant over all resistors, meaning V₁ = V₂ = 6 V.

The current gets divided to both resistors, they again have the same value, so the current gets divided equally, I₁ = I₂ = 1.5 A.

Again, if this was not the case you could always calculate using Ohm's law, I_i = V_p/R_i

(for both this is 6 V / 4 Ω = 1.5 A).