What is the efficiency of the circuit?

[Baiatul122001]

Homework Statement

Two electromotive voltage sources E1 and E2 are connected in parallel, and the internal resistors r1 and r2, and at the terminals of the circuit, a resistor R is connected. Is the efficiency of the circuit?

Relevant Equations

E=(E1/r1+E2/r2)/(1/r1+1/r2)=(E1r2+E2r1)/(r1+r2)

1/r=1/r1+1/r2=(r1+r2)/r1r2=>r=r1r2/(r1+r2)

I=E/(R+r)=[(E1r2+E2r1)/(r1+r2)]/[R+r1r2/(r1+r2)]=(E1r2+E2r1)/[R(r1+r2)+r1r2]

P=RI^2=R(E1r2+E2r1)^2/[R(r1+r2)+r1r2]^2

P'=EI=(E1r2+E2r1)/(r1+r2)*(E1r2+E2r1)/[R(r1+r2)+r1r2]=(E1r2+E2r1)^2/[R(r1+r2)^2+r1r2(r1+r2)]

η=P/P'={R(E1r2+E2r1)^2/[R(r1+r2)+r1r2]^2}/{(E1r2+E2r1)^2/[R(r1+r2)^2+r1r2(r1+r2)]}=R[R(r1+r2)+r1r2(r1+r2)]/[R(r1+r2)+r1r2]^2

η=R[R(r1+r2)+r1r2(r1+r2)]/[R(r1+r2)+r1r2]^2

It does not come out as the final answer

The answer in the textbook's

 

[gneill]

When you simplify a circuit with a Thevenin equivalent, you "lose" the details of what happens to the original components that were simplified away.

Consider a typical Thevenin equivalent with a load $R$:

The current for this circuit is simply $I = \frac{E_{th}}{R + R_{th}}$

The power in the load: $P = I^2 R$
Total power: $P' = I^2 (R + R_{th})$

So the efficiency is $\eta = \frac{I^2 R}{I^2(R + R_{th})}$

As you can see, the $I^2$'s cancel and you're left without anything that accounts for the details of the original voltage sources.

You'll have to analyze the original circuit as-is if you want to retain the details of the sources and their internal resistances.

 

[Baiatul122001]

But the efficiency of the circuit does not depend on the electromotive voltage, is the answer wrong?

 

[gneill]

The answer given for the problem is not wrong. The efficiency, as determined by power delivered to the load compared to the total power consumed by the whole circuit, does depend on how the various EMF's are utilized.

Consider: Suppose that there happened to be a voltage source "inside" the portion of the circuit that you took the Thevenin equivalent that did nothing but push current through a localized resistor, contributing nothing to the load. It would be "burning" power inefficiently, but would have to be taken into consideration non the less.

 

[rude man]

Easiest way is to change the two batteries plus internal resistaces to their respective Norton equivalents. A real piece of cake that way.

(I suppose "efficiency" means what gneill says in post 4.)