Advertisements
Advertisements
Questions
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?
In a potentiometer experiment, a cell of emf 1.25 V gives a balance point at 35 cm mark of the wire. If this cell is replaced by another cell, the balance point is at 63 cm mark. Calculate the emf of the second cell.
Advertisements
Solution
Emf of the cell, E1 = 1.25 V
Balance point of the potentiometer, l1 = 35 cm
The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm
The balance condition is given by the relation,
`E_1/E_2 = I_1/I_2`
`E_2 = E_1 xx I_2/I_1`
= `1.25 xx 63/35`
= 2.25 V
Therefore, the emf of the second cell is 2.25 V.
RELATED QUESTIONS
Two cells of emfs 1.5 V and 2.0 V, having internal resistances 0.2 Ω and 0.3 Ω, respectively, are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.
Distinguish between emf and terminal voltage of a cell.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
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?
Nichrome and copper wires of same length and same radius are connected in series. Current I is passed through them. Which wire gets heated up more? Justify your answer.
A long straight current carrying wire passes normally through the centre of circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify.
A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Ω as shown in the figure. Find the value of current in the circuit.
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 ?
Two non-ideal batteries are connected in series. Consider the following statements:-
(A) The equivalent emf is larger than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances.
Find the value of i1/i2 in the following figure if (a) R = 0.1 Ω (b) R = 1 Ω and (c) R = 10 Ω. Note from your answers that in order to get more current from a combination of two batteries, they should be joined in parallel if the external resistance is small and in series if the external resistance is large, compared to the internal resistance.
Consider N = n1n2 identical cells, each of emf ε and internal resistance r. Suppose n1 cells are joined in series to form a line and n2 such lines are connected in parallel.
The combination drives a current in an external resistance R. (a) Find the current in the external resistance. (b) Assuming that n1 and n2 can be continuously varied, find the relation between n1, n2, R and r for which the current in R is maximum.
Answer the following question.
A cell of emf E and internal resistance r is connected across a variable resistor R. Plot the shape of graphs showing a variation of terminal voltage V with (i) R and (ii) circuit current I.
Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.
A battery of EMF 10V sets up a current of 1A when connected across a resistor of 8Ω. If the resistor is shunted by another 8Ω resistor, what would be the current in the circuit? (in A)
Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor R.
Obtain expressions for (i) current flowing in the circuit, and (ii) the terminal potential differences across the equivalent cell.
The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is:
