Advertisements
Advertisements
प्रश्न
In a potentiometer experiment, balancing length is found to be 120 cm for a cell E1 of emf 2V. What will be the balancing length for another cell E2 of emf 1.5V? (No other changes are made in the experiment.)
Advertisements
उत्तर
``E_1/E_2 = l_1/l_2`
`2/1.5 = 120/l_2`
`l_2 = 90 cm`
APPEARS IN
संबंधित प्रश्न
State the principle of a potentiometer. Define potential gradient. Obtain an expression for potential gradient in terms of resistivity of the potentiometer wire.
Figure shows a long potentiometer wire AB having a constant potential gradient. The null points for the two primary cells of emfs ε1 and ε2 connected in the manner shown are obtained at a distance of l1 = 120 cm and l2 = 300 cm from the end A. Determine (i) ε1/ε2 and (ii) position of null point for the cell ε1 only.

When the balance point is obtained in the potentiometer, a current is drawn from ______.
A potentiometer wire has a length of 1.5 m and a resistance of 10 Ω. It is connected in series with the cell of emf 4 Volt and internal resistance 5 Ω. Calculate the potential drop per centimeter of the wire.
When the null point is obtained in the potentiometer, the current is drawn from the ______
In a potentiometer circuit, a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
If you are provided a set of resistances 2Ω, 4Ω, 6Ω and 8Ω. Connect these resistances so as to obtain an equivalent resistance of `46/3`Ω.
A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ε and internal resistance r. A cell C having emt `ε/2` and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in the figure shows no deflection is ______.
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.
