# Parallel circuits and power

#### barrett50

Joined Feb 1, 2016
14
*PDF relevant*

So, I have a parallel circuit with a voltage source and two resistors. My main question is, would the power supplied to X be changed in any way if R1 was not present?

For example if

V = 12 V

R1 = 5 kOhms

R2a = 2 kOhms

V=iR then the power supplied to R2a would be
12=i(2*10^3)
= 0,006 Amp

is this correct or would I have to consider R1 in this case?

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#### shteii01

Joined Feb 19, 2010
4,644
Parallel circuits are also known as Current Divider.

Your V produces a current, that current is then divided between the parallel branches. If you change a branch, then you change the current in both branches, and in turn change the power in both branches.

#### barrett50

Joined Feb 1, 2016
14
Parallel circuits are also known as Current Divider.

Your V produces a current, that current is then divided between the parallel branches. If you change a branch, then you change the current in both branches, and in turn change the power in both branches.
So this means I wouldn't be able to calculate Ohm's law on both of the resistors independently because the power that they receive are dependent on one another?

#### djsfantasi

Joined Apr 11, 2010
9,080
First, without X, how much current is your source providing? If X's resistance was 2*R1, how much current is your source providing?

#### barrett50

Joined Feb 1, 2016
14
First, without X, how much current is your source providing? If X's resistance was 2*R1, how much current is your source providing?
if X wasn't there then I = 0.0024 Amps (with R1 = 5*10^3 ohms)

#### shteii01

Joined Feb 19, 2010
4,644
So this means I wouldn't be able to calculate Ohm's law on both of the resistors independently because the power that they receive are dependent on one another?
Remember the current divider thing?
Your V generates current I. Current I reaches the parallel branches and is split into I1 and I2. This means that:
I=I1+I2

Now look at the power.
The power produced by voltage source is:
P=VI
But what is I? I=I1+I2
So the power is P=VI=V(I1+I2)=V(I1)+V(I2)
What is V(I1)? It is power dissipated by the resistor in a branch through which I1 is passing, let us call this power P1.
What is V(I2)? It is power dissipated by the resistor in a branch through which I2 is passing, let us call this power P2.
Therefore we can say that P=VI=V(I1+I2)=V(I1)+V(I2)=P1+P2

• barrett50

#### djsfantasi

Joined Apr 11, 2010
9,080
And the second half of my question?

#### barrett50

Joined Feb 1, 2016
14
Remember the current divider thing?
Your V generates current I. Current I reaches the parallel branches and is split into I1 and I2. This means that:
I=I1+I2

Now look at the power.
The power produced by voltage source is:
P=VI
But what is I? I=I1+I2
So the power is P=VI=V(I1+I2)=V(I1)+V(I2)
What is V(I1)? It is power dissipated by the resistor in a branch through which I1 is passing, let us call this power P1.
What is V(I2)? It is power dissipated by the resistor in a branch through which I2 is passing, let us call this power P2.
Therefore we can say that P=VI=V(I1+I2)=V(I1)+V(I2)=P1+P2
Ahhhhhhh. That makes so much sense. I can't believe I couldn't see that before. Thank you!

#### barrett50

Joined Feb 1, 2016
14
And the second half of my question?
I1 = 0.0024 (A)
I2 = 0.006 (A)

which means I = 0.0084 (A)

#### djsfantasi

Joined Apr 11, 2010
9,080
So now you know V in both branches. And you have to know I1 and I2... Then you can calculate P for both branches and, via Ohms Law, you can calculate the resistance presented by X.

• barrett50

#### crutschow

Joined Mar 14, 2008
32,910
Parallel circuits are also known as Current Divider.

Your V produces a current, that current is then divided between the parallel branches. If you change a branch, then you change the current in both branches, and in turn change the power in both branches.
Not true.
If you have a voltage source applied to a parallel branches, then a change in one branch only affects the power in the changed branch.
What you say if true only if you have a current source.

#### barrett50

Joined Feb 1, 2016
14
If i were to get I1 = 0.0024 (A) and I2 = -.003 (A) would I still be able to say that I1+I2 = I?

#### shteii01

Joined Feb 19, 2010
4,644
If i were to get I1 = 0.0024 (A) and I2 = -.003 (A) would I still be able to say that I1+I2 = I?
Remember, I1 and I2 are not magic. They are branch currents. Like in the tree, before you have branches, you have the main trunk. I is the current in the main trunk. When the main trunk splits into branches, then I splits into branch currents.

In this specific case we have 3 pieces, I, I1, I2. If you know any two pieces, you can then find the third.
- If you know I1 and I2, you can find I.
- If you know I and I1, you can find I2.
- If you know I and I2, you can find I1.

• barrett50

#### WBahn

Joined Mar 31, 2012
29,165
To reinforce what crutschow said (and which is the point of the hint given in the problem), the voltage across all elements that are in parallel is the same. If you have a voltage source as one of those elements, then that voltage source is defining the voltage across all of the elements. If you know the voltage across R1 and you know the value of R1, then you know the current in R1. The value of X does not enter into that at all (unless the value of X is zero in which case you have a mathematical contradiction that cannot be physically realized).