# Current Relationship

Discussion in 'Homework Help' started by tquiva, Dec 16, 2010.

1. ### tquiva Thread Starter Member

Oct 19, 2010
176
1
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?

2. ### tyblu Member

Nov 29, 2010
199
16
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).

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3. ### tquiva Thread Starter Member

Oct 19, 2010
176
1
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?

4. ### tquiva Thread Starter Member

Oct 19, 2010
176
1
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?

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,393
1,211
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.

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6. ### tyblu Member

Nov 29, 2010
199
16
$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
$

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