Advertisements
Advertisements
Question
In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.
Advertisements
Solution
Internal resistance of the cell, `r = R(l_1/l_2 - 1) = 9(350/300 -1) = 1.5` Ω
APPEARS IN
RELATED QUESTIONS
A cell of emf 'E' and internal resistance 'r' is connected across a variable resistor 'R'. Plot a graph showing variation of terminal voltage 'V' of the cell versus the current 'I'. Using the plot, show how the emf of the cell and its internal resistance can be determined.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Six lead-acid types of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
A resistor R is connected to a cell of-emf e and internal resistance r. The potential difference across the resistor R is found to be V. State the relation between e, V, Rand r.
A cell of emf ‘E’ and internal resistance ‘r’ draws a current ‘I’. Write the relation between terminal voltage ‘V’ in terms of E, I and r ?
A cell having an emf E and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by ______.
Five cells each of emf E and internal resistance r send the same amount of current through an external resistance R whether the cells are connected in parallel or in series. Then the ratio `("R"/"r")` is:
Two batteries of emf ε1 and ε2 (ε2 > ε1) and internal resistances r1 and r2 respectively are connected in parallel as shown in figure.

