# Current Relationship

#### tquiva

Joined Oct 19, 2010
176
Problem:

I first began by denoting the very top middle node as A.
At node A, I applied KCL:

Σi = I - I_R - I_L
=> I = I_R + I_L

So does this mean that I_R = I_L ? And the Answer is i?

#### tyblu

Joined Nov 29, 2010
199
How did you get $$I_R = I_L$$ from $$I = I_R + I_L$$? You need to define $$I_R$$ and $$I_L$$ in terms of $$R$$ using Ohm's law, $$V = I \times R$$. Just pretend the bottom terminal is ground (0V).

#### tquiva

Joined Oct 19, 2010
176
With the top node as A and bottom node a B, I = I_R + I_L simplifies to:
I = (V_A - V_B) / (2R) + (V_A - V_B) / (R)
= (V_A- V_B + 2V_A - 2V_B) / (2R)
= 3V_A - 3V_B / (2R)

Where would I go from here?

#### tquiva

Joined Oct 19, 2010
176
I also tried current division:

I_R = [(2/3)R]/[2R] * I_total = V_AB / (2R)

I_L = [(2/3)R]/[R] * I_total= V_AB / R

So I_R > I_L

Is this correct?

#### JoeJester

Joined Apr 26, 2005
4,266
I_L > I_R

It's a parallel circuit with three equal R's. Two of which are in one leg. I_L is .667 of I leaving I_R at .333 of I.

#### tyblu

Joined Nov 29, 2010
199
With the top node as A and bottom node a B...
$$I_R = \frac{V_A-V_B}{2R}\\ I_L = \frac{V_A-V_B}{R}\\ \frac{I_L}{I_R} = \frac{\left(V_A-V_B\right)/R}{\left(V_A-V_B\right)/2R} = \frac{2R}{R} = 2$$
$$I_L$$ is 2x larger than $$I_R$$.

Another way to look at this is that current is inversely proportional to resistance, or $$I \propto \frac{1}{R}$$. This gives:
$$I_L \propto 1/R \\ I_R \propto 1/2R \\ \frac{I_L}{I_R} \propto 2R/R = 2$$