Answer to Question #208285 in Classical Mechanics for Aysu

Question

An element is first connected across an external resistance R₁=2 Ω, and

then across an external resistance R₂= 0.5 Ω. Find the e.m.f. of the

element and its internal resistance if in each of these cases the power

evolved in the external circuit is the same and equal to 2.54 W.

Answer 1

$I= \frac{E}{R+r}$

$P=I^2R= \frac{E^2R}{(R+r)^2}$

$P_1=\frac{E^2*2}{(2+r)^2}$

$P_2=\frac{E^2*0.5}{(0.5+r)^2}$

$P_1 = P_2=2.54$

$\frac{E^2*2}{(2+r)^2}=\frac{E^2*0.5}{(0.5+r)^2}$

$4*(0.5+r)^2=(2+r)^2$

$2*(0.5+r)=(2+r)$

$r=1$

$P_1=I_1^2R_1$

$I_1=\sqrt{\frac{P_1}{R_1}}=\sqrt{\frac{2.54}{2}}=1.17$

$I_1= \frac{E}{R_1+r}$

$E=I_1(R_1+r)=1.17(2+1)=3.51$

$\text{Answer: }e.m.f = 3.51V;r=1\Omega$