- #1

Luchekv

- 66

- 1

First I will explain what I taught - Let us look at the following two circuits:

Specs for both:

10V Battery

The below values were calculated by the simulator (which is the source of confusion):

Circuit A:

Vd @ R1 = 10V - 10ma

Vd @ R2 = 10V - 5ma

Vd @ R3 = 10V - 3.333ma

Circuit B:

Vd @ R1 = 1.667V - 1.667ma

Vd @ R2 = 3.333V - 1.667ma

Vd @ R3 = 5.000V - 1.667ma

I used the analogy that voltage is like water pressure. Therefore in circuit A, each resistor will have the same pressure

**applied**to it.

Naturally, depending on the pipe thickness (resistor value) it will allow more or less water (charges) through. This

**difference**in charge is considered the

**voltage drop**.

Now in the above circuit A it shows that 10V is 'dropped' across each resistor. My sons confusion here is, if the whole '10V of pressure' is used to push water through pipe R1. Then how can there still be any pressure left for the other two pipes? Why is the simulation showing all 10V 'dropped' on all 3 resistors.

Even in other online examples the same principle is taught:

It would be more logical to me if this was explained in the following method for circuit A:

10V is

**applied**to each resistor - Out of those 10V:

R1 will require 1.667V of energy (work to push charges through)

R2 will require 3.333V of energy (work to push charges through)

R3 will require 5.000V of energy (work to push charges through)

Which equals the supplied 10V

Is this logic correct? I'm doubting myself due to lack of examples supporting this kind of explanation.