Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
Find the circuit in the three resistors shown in the figure.
Twelve wires, each of equal resistance r, are joined to form a cube, as shown in the figure. Find the equivalent resistance between the diagonally-opposite points a and f.
Solve the following question.
Using Kirchhoff’s rules, calculate the current through the 40 Ω and 20 Ω resistors in the following circuit.
Twelve wires each having a resistance of 3 Ω are connected to form a cubical network. A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of this network. Determine its equivalent resistance and the current along each edge of the cube.
State the principle of potentiometer.
The Kirchhoff's second law (ΣiR = ΣE), where the symbols have their usual meanings, is based on ______.
The e.m.f of The battery in a thermocouple is doubled. The rate of heat generated at one of the junction will.
The circuit in figure shows two cells connected in opposition to each other. Cell E1 is of emf 6V and internal resistance 2Ω; the cell E2 is of emf 4V and internal resistance 8Ω. Find the potential difference between the points A and B.
Two cells of voltage 10V and 2V and internal resistances 10Ω and 5Ω respectively, are connected in parallel with the positive end of 10V battery connected to negative pole of 2V battery (Figure). Find the effective voltage and effective resistance of the combination.
