Saturday, 16 March 2013

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:

Rl: stands for the load resistance
Rs: stands for the internal resistance of battery A
Rb: stands for the internal resistance of battery B
Vs: stands for the potential of the source of battery A
Vb: stands for the potential of the source of of battery B

Let's begin!!!
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: ∑ V ΑΒΓΔ = 0 ⇒ 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: ∑ V ΑΒΔΕ = 0 ⇒ 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 = 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 = I Γ R l − V s R s ( 5 ) Now from equation (1) we have: I Γ = Ι B − I A ⇒ ( using ( 4 ) ∧ ( 5 ) ) ⇒ I Γ = V b − I Γ R l R b − I Γ R l − V s R s ⇒ I Γ = V b R s + V s R b R b R s + R b R l + R s R l ⇒ (now multiplying an diverting with R s we are given) ⇒ I Γ = V b R s + V s R b R s R b R s + R b R l + R s R l R s = V b + V s R b R s R b + R b R s R l + R l = V b + V s R b R s R b + ( 1 + R b R s ) R l So the potential of the branch ΖΓ is: "Using Kirchhoff's law we have:" newline newline newline ~I_{Β} = I_{Α} + I_{Γ} ~~(1) newline newline newline "Ignoring the existance of the branch ΕΔ, For the closed circuit ΑΒΓΖ Kirchhoff's law says that:" newline newline newline sum V_{ΑΒΓΔ} = 0~~ drarrow ~~V_{b} - I_{Β}R_{b} - I_{Α}R_{s} - V_{s}~ = ~0 ~~~(2) newline newline newline "Ignoring the existance of the branch ΖΓ, For the closed circuit ΑΒΔΕ Kirchoff's law says that:" newline newline newline sum V_{ΑΒΔΕ} = 0~~ drarrow ~~V_{b} - I_{Β}R_{b} - I_{Γ}R_{l}~ = ~0 ~~~(3) newline newline newline "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)" newline newline newline newline "Solving equation (3) for " ~I_{Β}" gives us:" newline newline newline newline I_{B} ~= ~{V_{b} - I_{Γ}R_{l}} over {R_{b}} ~~~(4) newline newline newline newline "Substituting" ~I_{Β} "with (4) in equation (2) and solving for" ~I_{Α} " gives us:" newline newline newline newline I_{Α} ~= ~{I_{Γ}R_{l} - V_{s}} over {R_{s}} ~~~(5) newline newline newline "Now from equation (1) we have:" newline newline I_{Γ}~ =~ Ι_{B} - I_{A} ~~drarrow (using (4)&(5))drarrow ~~ I_{Γ}~ =~ ~{V_{b} - I_{Γ}R_{l}} over {R_{b}} ~ - ~ {I_{Γ}R_{l} - V_{s}} over {R_{s}} ~~drarrow ~~ newline newline newline I_{Γ}~ =~ ~{V_{b}R_{s} + V_{s}R_{b}} over {R_{b}R_{s}+R_{b}R_{l}+R_{s}R_{l}} ~~drarrow "(now multiplying an diverting with " ~R_{s}~ " we are given)"drarrow ~~ newline newline I_{Γ}~ =~ ~{{V_{b}R_{s} + V_{s}R_{b}}over{R_{s}}} over {{R_{b}R_{s}+R_{b}R_{l}+R_{s}R_{l}}over{R_{s}}} ~ = ~ {V_{b} + V_{s}{R_{b}}over{R_{s}}}over{R_{b} + {R_{b}}over{R_{s}}R_{l} + R_{l}} ~= ~ {V_{b} + V_{s}{R_{b}}over{R_{s}}}over{R_{b} + (1 + {R_{b}}over{R_{s}})R_{l}} newline newline newline "So the potential of the branch ΖΓ is:"
V ΖΓ = V ΕΔ = Ι Γ R l = V b + V s R b R s R b + ( 1 + R b R s ) R l ∗ R l newline newline V_{ΖΓ} ~=~ V_{ΕΔ} ~=~ Ι_{Γ}R_{l} ~=~ {V_{b} + V_{s}{R_{b}}over{R_{s}}}over{R_{b} + (1 + {R_{b}}over{R_{s}})R_{l}}~*~R_{l} newline newline

So now, having known the formula for estimating the potential V ΖΓ we must examine how it performs when changing the ratio 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 → 0 V ΖΓ = lim R b → 0 V b + V s R b R s R b + ( 1 + 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 → 0 V ΖΓ = lim R b → 0 V b + V s R b R s R b + ( 1 + R b R s ) R l ∗ R l = lim R b → 0 V b + V s R b + 2R l ∗ R l = 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 → 0 V ΖΓ = lim R s → 0 V b + V s R b R s R b + ( 1 + R b R s ) R l ∗ R l = lim R s → 0 V b R s R b + V s R s + ( 1 + 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! "So now, having known the formula for estimating the potential " ~V_{ΖΓ}~ " we must examine how it performs when changing the ratio " newline newline ~~~{R_{b}}over{R_{s}} newline newline "that is the relation between these two resistances. So now LET'S begin by assuming that " newline newline ~V_{b}">"V_{s} newline newline "meaning that source B provides more power than source A."~~~~~~~ newline newline newline newline "(Α) "V_{b}~>~V_{s}" (source B is stronger than source A):" newline newline " i) Given that "V_{b}~>~V_{s}", assume that" R_{b} " takes values close to zero so we have: " newline newline lim_{R_{b} toward 0} V_{ΖΓ} ~= ~ lim_{R_{b} toward 0} ~{V_{b} + V_{s}{R_{b}}over{R_{s}}}over{R_{b} + (1 + {R_{b}}over{R_{s}})R_{l}}~*~R_{l} ~=~ V_{b} newline newline " which means that IF the larger battery(battery B) has an internal resistance is so small (relative to the internal" newline " resistance of the smaller battery(battery A)) that could be neglected, THEN the combined supplying voltage" newline " of the two batteries tends to be equal to the voltage of the larger battery" newline V_{b} "." newline newline newline newline " ii) Given that " ~V_{b}~>~V_{s}", assume that" ~R_{b} ~=~ R_{s} " and " R_{b} " takes values close to zero so we have:" newline newline lim_{R_{b} toward 0} V_{ΖΓ} ~= ~ lim_{R_{b} toward 0} ~{V_{b} + V_{s}{R_{b}}over{R_{s}}}over{R_{b} + (1 + {R_{b}}over{R_{s}})R_{l}}~*~R_{l} ~=~ lim_{R_{b} toward 0} ~{V_{b} + V_{s}}over{R_{b} + 2R_{l}}~*~R_{l} ~=~ {V_{b} + V_{s}}over{2} newline newline " which means that IF the batteries have equal internal resistances AND so small that could be neglected, " newline " 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." newline newline newline newline " 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} newline " performs in such a case. So we have:" newline newline lim_{R_{s} toward 0} V_{ΖΓ} ~= ~ lim_{R_{s} toward 0} ~{V_{b} + V_{s}{R_{b}}over{R_{s}}}over{R_{b} + (1 + {R_{b}}over{R_{s}})R_{l}}~*~R_{l} ~=~ lim_{R_{s} toward 0} ~{V_{b}{R_{s}}over{R_{b}} + V_{s}}over{R_{s} + (1 + {R_{s}}over{R_{b}})R_{l}}~*~R_{l} ~=~ V_{s} newline newline " which means that IF the smaller battery(battery A) has an internal resistance is so small (relative to the internal " newline " resistance of the larger battery(battery B)) that could be neglected, THEN the combined supplying " newline " voltage of the two batteries tends to be equal to the voltage of the smaller battery" newline V_{s} "." newline newline newline newline "(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 " newline " i will give it to you as a homework. :P " newline " I HOPE YOU ENJOYED IT!!! PEACE! Jesus Christ Bless You all!"
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