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.
Advertisements
उत्तर
r = `R (I_1/I_2 -1)`
= `10 (110/100 -1)`
= 1 Ω
APPEARS IN
संबंधित प्रश्न
Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?
The net resistance of an ammeter should be small to ensure that _______________ .
The net resistance of a voltmeter should be large to ensure that ______________ .
State the uses of a potentiometer.
What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is decreased?
Describe how a potentiometer is used to compare the emf's of two cells by the combination method.
The emf of a cell is balanced by a length of 120 cm of a potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.
To determine the internal resistance of a cell by using potentiometer, the null point is at 1 m when cell is shunted by 3 Ω resistance and at a length 1.5 m when cell is shunted by 6 Ω resistance. The internal resistance of the cell is ______.
When two cells of e.m.f 1.5 V and 1.1 V connected in series are balanced on a potentiometer, the balancing length is 260 cm. The balancing length, when they are connected in opposition is (in cm) ____________.
In the potentiometer experiment, the balancing length with cell E1 of unknown e.m.f. is ℓ1 cm. By shunting the cell E1 with resistance 'R' which is equal to internal resistance (r) of the cell E1, the balancing length ℓ2 is ______
