Question

The circuit shown in Figure extends to the right into infinity. Each battery has the electromotive force $V$(unknown) and the internal resistance $r$ (known). Each resistor has the resistance $4r$. The reading of the ideal ammeter shown in the diagram is $I$. Find the value of $V$ in terms of $I$ and $r$.

Answer 1

Let net resistance and net EMF across CD and be R and E.Then net resistance and net EMF will also be R and E. In this way infinite series will continue.

$\frac{1}{R}=\frac{1}{r}+\frac{1}{(R+4r)}R+4r=\frac{4r^2}{R}\to 1\epsilon =\frac{\frac{\epsilon }{R+4r}+\frac{V}{R}}{\frac{1}{R+4r}+\frac{1}{r}}\epsilon (R+4R)=v(R+4R)\epsilon =V$

Also I= $I=\frac{\epsilon R}{4r^2}=\frac{4R}{4r^2}R=\frac{4r^{2}I}{V}\to 3$

From (1) and (3)

$\frac{4r^{2}I}{V}+4r=\frac{4r^{2}V}{{4r}^{2}I}=\frac{V}{I}4r^{2}I+4rV=\frac{V2}{I}{V}^2-4rIV-4r^2{I}^2=0V=2Ir(1+\sqrt{2})$