Question

In the circuit shown aside, the resistance of a uniform wire $AB$ is $R_o$. The ammeter readings are noted by sliding the contact $C$ along the length $AB$ of the wire. If the maximum current through ammeter is equal to twice the minimum current, then,

• A. $R_o=r$
• B. $R_o=2r$
• C. $R_o=3r$
• D. $R_o=4r$

Answer 1

The correct option is D $R_o=4r$

Let fR${}_o$ be the resistance of AC and (1 - f) R${}_o$ be that of BC, where $0\le f\le 1$.

Now, applying Kirchhoff voltage law in both loops we get,

$E-I_{1}f{R}_o-Ir=0$ ...................................... (1)

$E-I_2(1-f)R_o-Ir=0$ ........................................ (2)

Also, $I_1+I_2=I$

Performing $Eq\left(1\right)\times (1-f)+Eq\left(2\right)\times f$ we get

$E(1-f+f)-f(1-f)R_o(I_1+I_2)-Ir(1-f+f)=0$

$\therefore I[r+f(1-f\left)R_o\right]=E$

or $I=\frac{E}{[r+f(1-f\left)R_o\right]}$ ................ (3)

Now, I is maximum or minimum when denominator of right-hand side of equation (3) is minimum or maximum respectively.

$\therefore$ I is maximum, when f = 0 or f = 1 and

I is minimum, when $\frac{d}{df}[r+f(1-f\left)R_o\right]=0\Rightarrow (1-2f)=0$

$\Rightarrow f=\frac{1}{2}$

Thus $I_{max}=\frac{E}{r},I_{min}=\frac{E}{[r+\frac{R_o}{4}]}$

$\therefore \frac{I_{max}}{I_{min}}=\frac{r+\frac{R_o}{4}}{r}=[1+\frac{R_o}{4r}]$

Hence, $I_{max}$ = 2 $I_{min}$ if R${}_o$ = 4r