**Disclaimer:**This is

__a homework question.__

**NOT****Situation:**This is a personal project where I am trying to model a split rail configured circuit with two

__constant power__loads, Pp and Pn, with a "shared" common return to ground. Reistance, R, is equal across all lines. Pp does not necessarily equal the magnitude of Pn.

**Issue**: The fact there are two constant power loads makes it difficult to extract an expression for I+ or I- based on the change in power consumed by the loads. I have so far developed an expression for the current as a function of the current in the opposite leg (ie

*I+ as a function of I- and vice versa*), but I need to isolate purely I+ without any coupled dependancy.

**Work done so far...**

Note:I+ and I- are the currents sourced by the positive rail and negative rail respectively. These quantities are henceforth referenced as I1 and I2 respectively.

Note:

**Knowns**: V, R, Pp, Pn

1. I began by developing an expression for I1 and I2 as a fucntion of the voltage drop across the load.

\( I_1 = \frac{1}{R}\left(V - \frac{2}{3}V_{pos} - \frac{1}{3}|V_{neg}|\right) \)

\( I_2 = -\frac{1}{R}\left(V - \frac{2}{3}|V_{neg}| - \frac{1}{3}V_{pos}\right) \)

2. Next I put Vpos and Vneg in terms of the power and their respective current through the load.\( I_2 = -\frac{1}{R}\left(V - \frac{2}{3}|V_{neg}| - \frac{1}{3}V_{pos}\right) \)

\( P_p := V_{pos}I_1\)

\( P_n := |V_{neg}|I_2=NP_p=N V_{pos}I_1\)

3. Substituting into the expressions for I1 and I2 reduces to\( P_n := |V_{neg}|I_2=NP_p=N V_{pos}I_1\)

*\( N = \frac{P_n}{P_p}\)*

wherewhere

\( I_1 = -\frac{P_p}{3R}\left(\frac{I_2}{(I_2)^2 - \frac{V}{R}I_2 + \frac{2N}{3R}P_p}\right) \)

\( I_2 = \frac{NP_p}{3R}\left(\frac{I_1}{(I_1)^2 - \frac{V}{R}I_1 + \frac{2}{3R}P_p}\right) \)

\( I_2 = \frac{NP_p}{3R}\left(\frac{I_1}{(I_1)^2 - \frac{V}{R}I_1 + \frac{2}{3R}P_p}\right) \)

4. From here on, I'm attempting to isolate I1 by considering the voltage drop across one mesh.

\( V=I_1R + \frac{P_p}{I_1} + (I_1 - I_2) \)

5. Which, when I1 from step 3 is substituted into the mesh equation in step 4, the following behemoth is found:

\( I_1 = \frac{V(I_1)^2\left(3R(I_1)^2 - 3I_1V+2P_p\right)}{6(I_1)^4R^2 - 6RV(I_1)^3 +(I_1)^2(7P_pR+NP_pR)-3P_pVI_1 + 2(P_p)^2} \)

**Comments moving forward**: Provided step 5, I have clearly hit a snag. I am not sure how best to proceed to isolate the current I1. I considered a Taylor expansion (or some variant) in the denominator and bound the expression, but this may be unnecessarily tedious. Unfortunately, any other manual approach seems equally tedious.

Any help would be greatly appreciated!