One of the circuits for the measurement of resistance by potentiometer is shown. The galvanometer is connected at point `A` and zero deflection is observed is observed at length `PJ = 30 cm`. In second case the secondary cell is changed.

Take `E_(s) = 10 V` and `r = 1 Omega` in `1^(st)` reading and `E_(s) = 5 V` and `r = 2 Omega` in `2^(nd)` reading. In second case, the zero deflection is observed at length `PJ = 10 cm`. What is the resistance `R` (in ohm)?

Options

A. `1 Omega`
B. `2 Omega`
C. `3 Omega`
D. `4 Omega`

Answer

A

Explanation

(a) When there is no current through galvanometer, current throuhg `R` is
`I = (E_(S))/(R + r)`
Potential difference `= IR = (E_(S) R)/(R + r)`
Now `(E_(S) R)/(R + r) = K l`
For first case, `(10 R)/(R + 1) = K (30)`
Divide (i) and (ii) to get `R = 10 Omega`