HellaScience: Combining two different voltage sources in parallel

*NOTE: If the formulas are not rendered properly try open the page using mozilla firefox.

Assume the following circuit:

Assume THAT:

R_l: stands for the load resistance
R_s: stands for the internal resistance of battery A
R_b: stands for the internal resistance of battery B
V_s: stands for the potential of the source of battery A
V_b: stands for the potential of the source of of battery B

Let's begin!!!

\begin{matrix} Using Kirchhoff's law we have: \\ I_{Β} = I_{Α} + I_{Γ} (1) \\ Ignoring the existance of the branch ΕΔ, For the closed circuit ΑΒΓΖ Kirchhoff's  law says that: \\ \sum V_{ΑΒΓΔ} = 0 \Rightarrow V_{b} - I_{Β} R_{b} - I_{Α} R_{s} - V_{s} = 0 (2) \\ Ignoring the existance of the branch ΖΓ, For the closed circuit ΑΒΔΕ Kirchoff's law says that: \\ \sum V_{ΑΒΔΕ} = 0 \Rightarrow V_{b} - I_{Β} R_{b} - I_{Γ} R_{l} = 0 (3) \\ Now we have to solve a system of 3 equations (1,2 & 3) Let's start mathematics (μαθηματικά in Greek :P μ=m, θ = th, η=e, κ=c) \\ Solving equation (3) for I_{Β} gives us: \\ I_{B} = \frac{V_{b} - I_{Γ} R_{l}}{R_{b}} (4) \\ Substituting I_{Β} with (4) in equation (2) and solving for I_{Α} gives us: \\ I_{Α} = \frac{I_{Γ} R_{l} - V_{s}}{R_{s}} (5) \\ Now from equation (1) we have: \\ I_{Γ} = Ι_{B} - I_{A} \Rightarrow (using (4) \land (5)) \Rightarrow I_{Γ} = \frac{V_{b} - I_{Γ} R_{l}}{R_{b}} - \frac{I_{Γ} R_{l} - V_{s}}{R_{s}} \Rightarrow \\ I_{Γ} = \frac{V_{b} R_{s} + V_{s} R_{b}}{R_{b} R_{s} + R_{b} R_{l} + R_{s} R_{l}} \Rightarrow (now multiplying an diverting with R_{s} we are given) \Rightarrow \\ I_{Γ} = \frac{\frac{V_{b} R_{s} + V_{s} R_{b}}{R_{s}}}{\frac{R_{b} R_{s} + R_{b} R_{l} + R_{s} R_{l}}{R_{s}}} = \frac{V_{b} + V_{s} \frac{R_{b}}{R_{s}}}{R_{b} + \frac{R_{b}}{R_{s}} R_{l} + R_{l}} = \frac{V_{b} + V_{s} \frac{R_{b}}{R_{s}}}{R_{b} + (1 + \frac{R_{b}}{R_{s}}) R_{l}} \\ So the potential of the branch ΖΓ is: \end{matrix}

\begin{matrix} V_{ΖΓ} = V_{ΕΔ} = Ι_{Γ} R_{l} = \frac{V_{b} + V_{s} \frac{R_{b}}{R_{s}}}{R_{b} + (1 + \frac{R_{b}}{R_{s}}) R_{l}} * R_{l} \end{matrix}

\begin{matrix} So now, having known the formula for estimating the potential V_{ΖΓ} we must examine how it performs when changing the ratio \\ \frac{R_{b}}{R_{s}} \\ that is the relation between these two resistances. So now LET'S begin by assuming that \\ V_{b} > V_{s} \\ meaning that source B provides more power than source A. \\ (Α) V_{b} > V_{s} (source B is stronger than source A): \\ i) Given that V_{b} > V_{s}, assume that R_{b} takes values close to zero so we have: \\ \lim_{R_{b} \to 0} V_{ΖΓ} = \lim_{R_{b} \to 0} \frac{V_{b} + V_{s} \frac{R_{b}}{R_{s}}}{R_{b} + (1 + \frac{R_{b}}{R_{s}}) R_{l}} * R_{l} = V_{b} \\ which means that IF the larger battery(battery B) has an internal resistance is so small (relative to the internal \\ resistance of the smaller battery(battery A)) that could be neglected, THEN the combined supplying voltage \\ of the two batteries tends to be  equal to the voltage of the larger battery \\ V_{b} . \\ ii) Given that V_{b} > V_{s}, assume that R_{b} = R_{s} and R_{b} takes values close to zero so we have: \\ \lim_{R_{b} \to 0} V_{ΖΓ} = \lim_{R_{b} \to 0} \frac{V_{b} + V_{s} \frac{R_{b}}{R_{s}}}{R_{b} + (1 + \frac{R_{b}}{R_{s}}) R_{l}} * R_{l} = \lim_{R_{b} \to 0} \frac{V_{b} + V_{s}}{R_{b} + {2R}_{l}} * R_{l} = \frac{V_{b} + V_{s}}{2} \\ which means that IF the batteries have equal internal resistances AND so small that could be neglected, \\ THEN the combined supplying voltage of the two batteries tends to be equal to the sum of the voltages of these batteries divided by 2. \\ iii) Given that V_{b} > V_{s}, now it's time for R_{s} to take values close to zero so let's see how potential V_{ΖΓ} \\ performs in such a case. So we have: \\ \lim_{R_{s} \to 0} V_{ΖΓ} = \lim_{R_{s} \to 0} \frac{V_{b} + V_{s} \frac{R_{b}}{R_{s}}}{R_{b} + (1 + \frac{R_{b}}{R_{s}}) R_{l}} * R_{l} = \lim_{R_{s} \to 0} \frac{V_{b} \frac{R_{s}}{R_{b}} + V_{s}}{R_{s} + (1 + \frac{R_{s}}{R_{b}}) R_{l}} * R_{l} = V_{s} \\ which means that IF the smaller battery(battery A) has an internal resistance is so small (relative to the internal \\ resistance of the larger battery(battery B)) that could be neglected, THEN the combined supplying \\ voltage of the two batteries tends to be  equal to the voltage of the smaller battery \\ V_{s} . \\ (B) in the same way we work for the case where we assume that V_{s} > V_{b} (source A is stronger than source B):  ... B U T \\ i will give it to you as a homework. :P \\ I HOPE YOU ENJOYED IT!!!  PEACE! Jesus Christ Bless You all! \end{matrix}

2 comments:

Valour9 April 2014 at 06:51
nice nice good stuff
brucer51 September 2014 at 02:12
Great, your formulas appeared with FireFox, but not under Chrome or Opera.
I used your formulas with Vb < Vs,
I used the following to to calculation Vs =1.5V, Vb =1.2V (subscripts in lower-case)
Rs= 0.1 ohm Rb = 0.2 ohm Rt = 3 ohm (changed Rl to Rt, and Ir to It, 't' for top of circuit)
Solution: Vzr = 1.36956, It = .45652 Amps, Ib = -.0848 amps, Ia = -1.3043 amps
So, Ib = Ia + It >> -.0848 = -1.3043 + .4565
Vzr = It * Rt = .45652 * 3 = 1.36956 [THAT's the Blue Rectangle]

Re: the limits in ii), the first limit lim Rb → 0 evaluates to Vb as follows:

(Vb + 0) Vb * Rl
lim Rb → 0 ... = ----------------- * Rl = ---------- = Vb
0 +(1 + 0) Rl Rl

This result is analogous to the second limit
lim Rb → 0 ... = Vs

This means that the conclusion "the combined supplying voltage of the two batteries
tends to be equal to the voltage of the smaller battery" applies equally to each of the limits"

HOMEWORK: Assume Vs=Vb:
For the first limit lim Rb → 0 , substitute Vb for Vs >>> evaluates to Vb,
, substitute Vs for Vb >>> evaluates to Vs

For the second limit lim Rs → 0 , substitute Vs for Vb >>> evaluates to Vs.
, substitute Vb for Vs >>> evaluates to Vb.

Saturday, 16 March 2013

Combining two different voltage sources in parallel

2 comments: