A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key K3 is open. The balance point is at a length l1 cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm from P when key K3 is plugged in between 3 and 1. The ratio of two resistances (R1/R2), is found to be :

Question

A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key K3 is open. The balance point is at a length l1 cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm from P when key K3 is plugged in between 3 and 1. The ratio of two resistances (R1/R2), is found to be : (A) (l1/l1 + l2) (B) (l2/l2 - l1) (C) (l1/l1 - l2) (D) (l1/l2 - l1)

Tardigrade Admin · Accepted Answer

When key is at point V1 = iR1 = xl1 when key is at (3) V2 = i (R1 + R2) = xl2 (R1/R1+R2) = (l1/l2) (R1/R2) = (l1/l2-l1)

$\frac{l _{1}}{l _{1} + l _{2}}$

$\frac{l _{2}}{l _{2} - l _{1}}$

$\frac{l _{1}}{l _{1} - l _{2}}$

$\frac{l _{1}}{l _{2} - l _{1}}$

When key is at point
$V_{1} = i R_{1} = x l_{1}$
when key is at (3)
$V_{2} = i (R_{1} + R_{2}) = x l_{2}$
$\frac{R _{1}}{R _{1} + R _{2}} = \frac{l _{1}}{l _{2}}$
$\frac{R _{1}}{R _{2}} = \frac{l _{1}}{l _{2} - l _{1}}$

l1​+l2​l1​​

l2​−l1​l2​​

l1​−l2​l1​​

l2​−l1​l1​​

When key is at point V1​=iR1​=xl1​ when key is at (3) V2​=i(R1​+R2​)=xl2​ R1​+R2​R1​​=l2​l1​​ R2​R1​​=l2​−l1​l1​​

$\frac{l _{1}}{l _{1} + l _{2}}$

$\frac{l _{2}}{l _{2} - l _{1}}$

$\frac{l _{1}}{l _{1} - l _{2}}$

$\frac{l _{1}}{l _{2} - l _{1}}$

When key is at point
$V_{1} = i R_{1} = x l_{1}$
when key is at (3)
$V_{2} = i (R_{1} + R_{2}) = x l_{2}$
$\frac{R _{1}}{R _{1} + R _{2}} = \frac{l _{1}}{l _{2}}$
$\frac{R _{1}}{R _{2}} = \frac{l _{1}}{l _{2} - l _{1}}$