Advertisements
Advertisements
Question
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?
Advertisements
Solution
Emf of the cell1, ξ1 = 1.25 V
Balancing length of the cell, l1 = 35 cm = 35 x 10-2 m
Balancing length after interchanged, l2 = 63 cm = 63 x 10-2 m
Emf of the cell1, ξ2 = ?
The ration of emf’s, `ξ_1/ξ_2 = l_1/l_2`
The ration of emf’s, ξ2 = ξ1 = `(l_2/l_1)`
= 1. 25 `xx ((63 xx 10^-2)/(35 xx 10^-2)) = 12.5 xx 1.8`
ξ2 = 2.25 V.
APPEARS IN
RELATED QUESTIONS
Determine the current in each branch of the network shown in figure.
State Kirchhoff's rules and explain on what basis they are justified.
Determine the equivalent resistance of networks shown in Fig.
State Kirchhoff's rules for an electric network. Using Kirchhoff's rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.
Find the circuit in the three resistors shown in the figure.
Solve the following question.
Using Kirchhoff’s rules, calculate the current through the 40 Ω and 20 Ω resistors in the following circuit.
State Kirchhoff’s current rule.
The instrument for the accurate measurement of the e.m.f of a cell is ______.
What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate `R_(unknown)` by any other method?
Why are alloys used for making standard resistance coils?
