Advertisements
Advertisements
प्रश्न
In a potentiometer, a cell is balanced against 110 cm when the circuit is open. A cell is balanced at 100 cm when short-circuited through a resistance of 10 Ω. Find the internal resistance of the cell.
उत्तर
r = `R (I_1/I_2 -1)`
= `10 (110/100 -1)`
= 1 Ω
APPEARS IN
संबंधित प्रश्न
State the underlying principle of a potentiometer ?
The potentiometer wire AB shown in the figure is 40 cm long. Where should the free end of the galvanometer be connected on AB, so that the galvanometer may show zero deflection?
Draw a labelled circuit diagram of a potentiometer to compare emfs of two cells. Write the working formula (Derivation not required).
Define or describe a Potentiometer.
Find the equivalent resistance between the terminals of A and B in the network shown in the figure below given that the resistance of each resistor is 10 ohm.
Two cells having unknown emfs E1 and E2 (E1 > E2) are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell E2 is connected, so as to oppose cell E1, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells `("E"_1/"E"_2)` is ______.
AB is a wire of potentiometer with the increase in value of resistance R, the shift in the balance point J will be:
The instrument among the following which measures the e.m.f of a cell most accurately is ______
A particle carrying 8 electron charges starts from rest and is accelerated through a potential difference of 9000 V. Calculate the KE acquired by it in keV.
