**Current ElectricityÂ Important Questions for CBSE Class 12 Physics Potentiometer, Cell and their Combinations**

**1.Cell** A device to maintain a steady current in an electric circuit is electrolytic cell. It has two electrodes positive (F) and negative (N) as shown in figure below.

**2**.**Internal Resistance** The electrolyte through which a current flows has a finite resistance r, called the internal resistance.

**3.**The **emf of cell** is the potential difference between two terminals of cell when there is no flow of current through it.

The **potential difference** between two terminals of cell when current flows through it, is known as terminal voltage or terminal potential difference (V) of cell.

**13.**The sensitivity of a potentiometer refers to the capability of measuring very small potential difference and exhibit change in balancing length even on very small change in potential difference.

The sensitivity of potentiometer can be increased by increasing the number of wires of potentiometer and hence, decreasing the value of potential gradient.

**14.**The potentiometer works only when(i) the terminal voltage applied by driving cell is greater than the emf of primary cell. (ii) the positive terminals of driving cell and primary cell are connected at the zero end of potentiometer wire.

**15.**The potentiometer is a better device to measure potential difference than a voltmeter as null point method is used and hence, it can measure even the emf of cell but voltmeter cannot. It measures potential difference with greater accuracy.

**Previous YearÂ Examination Questions**

**1 Mark Questions**

**1.State the underlying principle of aÂ potentiometer? Â [Delhi 2014 c]**

**Ans**.The potentiometer works on the principle that potential difference across any two points of uniform current carrying conductor is directly proportional to the length between the two points.

**2.Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance What is the current through this resistance? [All India 2013]**

**Ans**.The cells are arranged as shown in the circuit

As the internal resistance of cells is negligible, so total resistance of the circuit = R

So. current through the resistance. **I=E/R**

(In parallel combination, potential is same as the single cell)

**Ans.**Since, the positive terminal of the batteries are connected together, so the equivalent emf of the batteries is given by Â£ = 200 -10 = 190 V

Hence, the current in the circuit is given by Â I=E/R=190/38=5 A

**4.The emf of a cell is always greaterÂ than its terminal voltage. Why? Give reason.Â Â Â Â Â Â Â Â Â [Delhi 2013]**

**Ans**.The emf of a cell is greater than its terminal Â voltage because there is some potential drop across the cell due to its small internal resistance

**5.A cell of emf E and internal resistance r draws a current Write the relation between terminal voltage V in terms of E, I and r.[Delhi 2013]**

**Ans.**When a current Idraws from a cell of emf EÂ and internal resistance r, then the terminal voltage is

V = E – Ir.

**6.A resistance R is connected across a cell of emf E and internal resistance r. Now, a potentiometer measures the potential difference between the terminals of the cells as V. Write the expression for r in terms of E, V and R.**

** **

**Ans.**

**7.A (i) series (ii) parallel combinationÂ of two given resistors is connected, one-by-one, across a cell. In which case, will the terminal potential difference across the cell have a higher value?[All India 2008 C]**

**Ans**.The equivalent resistance combination of resistances is (i) greater than the greatest resistance in series combination and (ii) smaller than the least value of resistance in parallel combination.

The terminal potential difference across the cell is higher inÂ series combination as V = E – IrÂ and due to higher resistance, current I is less in series combination.

**8.The plot of the variation of potential difference across a combination of three identical cells in series versus current is as shown in figure. What is the emf of each cell?**

**Ans.**Terminal potential difference across a cell can be obtained by subtracting potential drop across internal resistance of the cell from the emf of the cell.

v Terminal voltage across cell combination,

V = E – Ir

when current I=0

=> Â Â V = E

From graph, when I = 0, V = 6 V

=> Â Â emf E = 6 V

**2 Marks Questi0ns**

**9.A cell of emf E and internal resistance r is connected across a variable resistor Plot a graph showing variation of terminal voltage V of the cell versus the current I. Using the plot, show the emf of the cell and its internal resistance can be determined.[All India 2014]**

**Ans.**

**Ans.**

**Ans**.Refer to ans. 10. (Ans. 1.2V).

**Ans**.Refer to ans. 10. (Ans. 2.25 V).

**Ans.**

**14.Describe briefly with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.[All India 2013]**

**Ans.**

**15.Two students X and Y perform an experiment on potentiometer separately using the circuit given below**

** **

** Keeping other parameters unchanged, how will the position of the null point be affected if**

** (i)X increases the value of resistance R in the set up by keeping the key closed and the key K _{2} open?**

**(ii)Y decreases the value of resistance S in the set up, while the key K**

_{2}remains open and then K_{t}closed?**Justify your answer.[HOTS; Foreign 2012]**

**Ans.**

**16.Two cells of emf 2E and E and internal resistances 2r and r respectively, are connected in parallel. Obtain the expressions for the equivalent emf and the internal resistance of the combination.[All India 2010 C]**

**Ans.**

**17.Three cells of emf E,2E and 5 Eh aving internal resistances r, 2r and 3r, variable resistance R as shown in the figure. Find the expression for the current. Plot a graph for variation of current with R.**

** [All India 2010 C]**

** **

**Ans.**

**18.A cell of emf E and internal resistanceÂ r is connected across a variable resistor R. Plot a graph showing the variation of terminal potential V with resistance!?.[Delhi 2009]**

**Ans.**

**19.Plot a graph showing the variation ofÂ terminal potential difference across a cell of emf E and internal resistance r with current drawn from it. Using this graph, how does one determine the emf of the cell? Â [Delhi 2009 c]**

**Ans.**

**3 Marks Questions**

**Ans.**

**21.(i) State the underlying principle of aÂ potentiometer. Why is it necessary to (i) use a long wire, (ii) have uniform area of cross-section of the wire and (iii) use a driving cell whose emf is taken to be greater than the emfs of the primary cells?**

** (ii) In a potentiometer experiment, if the area of the cross-section of the wire increases uniformly from one end to the other, draw a graph showing how potential gradient would vary as the length of the wire increases from one end.[All India 2014 C]**

**Ans**.(i) Principle of PotentiometerÂ The potential drop across the length of a steady current carrying wire of uniform cross-section is proportional to the length of the wire.

(a)We use a long wire to have a lower value of potential gradient (i.e a lower “least count” or greater sensitivity of the potentiometer.

(b)The area of cross-section has to be uniform to get a ‘uniform wire’ as per the principle of the potentiometer.

(c)The emf of the driving cell has to beÂ greater than the emf of the primary cells as otherwise, no balance point would be obtained

**22.In the figure, a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs E _{x} and E_{2 }connected in the manner shown, are obtained at a distance of 120 cm and 300 cm from the end A**

**FindÂ (i) E1 / E**

_{2}and**(ii) position of null point for the cellÂ E1**

**How is the sensitivity of a potentiometer increased?Â [Foreign 2014; Delhi 2012]**

**Ans.**

**23.State the underlying principle of a potentiometer. Write two factors on which the sensitivity of a potentiometer depends.**

** **

** In the potentiometer circuit shown in the figure, the balance point is at X. State, giving reason, how the balance point is shifted when**

** (i)resistance R is increased**

** (ii)resistance S is increased, keeping R constant?[Compartment 2013]**

**Ans.**We use a long wire to have a lower value of potential gradient (i.e a lower “least count” or greater sensitivity of the potentiometer Â The two factors on which the sensitivity of a potentiometer depends are

(a)the value of potential gradient (K)

(b)by increasing the length of potentiometer wire. From the circuit diagram,

(i) if R is increased, the current through the potentiometer wire will decrease. Due to it, the potential gradient of potentiometer wire will also decrease. Thus, the position of **J** will shift towards **B.**

(ii) if S is increased, keeping R constant, the position of J will shift towards A.

**Ans.**

**25.With the help of circuit diagram,Â explain how a potentiometer can be used to compare emf of two primary cells?Â [Delhi 2011]**

**Ans.**

**26.State the underlying principle of aÂ potentiometer. Describe briefly, giving the necessary circuit diagram, how a potentiometer is used to measure the internal resistance of a given cell?[Foreign 2011]**

**Ans.**Principle of Potentiometer The potential difference across any two points of current carrying , wire, having uniform cross-sectional Â area andÂ material of the potentiometer is directlyÂ proportional to the length between the two points

The circuit diagram of potentiometer for determining internal resistance of a given cell is shown

Measurment of Internal Resistance of a Cell Using Potentiometer

**27.Two cells of emf E _{x}, E_{2} and internal resistances r_{a} and r_{2} respectively are connected in parallel as shown in the figure.**

**Deduce the expressions for**

**(i)the equivalent emf of the combination.**

**(ii)the equivalent resistance of the combination and**

**(iii)the potential difference between the points A and [Foreign 2010]**

**Ans.**

**28.Draw the circuit diagram of a potentiometer which can be used to determine the internal resistance r of a given cell of emf Explain briefly how the internal resistance of the cell is determined? [Delhi 2010,2008 C]**

**Ans.**

**29.Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.2 V. When the terminals of the cell are also connected to a resistance of 5 ohms as shown in the circuit, the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell.**

** **

**Ans.**The high resistance voltmeter means that no current with flow through it hence, there is no potential difference across it. So, the reading shown by the high resistance voltmeter can be taken as the emf of the cell.

The internal resistance of a cell depends on

(i) the concentration of electrolyte and

(ii) distance between the two electrodes

**30.(i)State the principle of working of a potentiometer.**

** (ii)Figure shows the circuit diagram of a potentiometer for determining the emf of E cell of negligible internal resistance.**

** **

** (a)What is the purpose of using high resistance ?**

** (b)How does the position of balance point (J) change when the resistance i? _{x} is decreased?**

**(c)Why cannot the balance point be obtained**

**when the emf E is greater than 2 V.****when the key (K) is closed?[Foreign 2009]**

**Ans**.(i) Principle of PotentiometerÂ The potential drop across the length of a steady current carrying wire of uniform cross-section is proportional to the length of the wire.

(a)We use a long wire to have a lower value of potential gradient (i.e a lower “least count” or greater sensitivity of the potentiometer.

(b)The area of cross-section has to be uniform to get a ‘uniform wire’ as per the principle of the potentiometer.

(c)The emf of the driving cell has to beÂ greater than the emf of the primary cells as otherwise, no balance point would be obtained

(ii) (a) To protect the galvanometer from flow of high current.

(b) Balance point, J shift towards **A**

(c)Potential drop across the cell cannot become equal to potential difference across any two point of the curve. Potential difference across cellÂ become zero

**31.A circuit using a potentiometer and battery of negligible internal resistance is set up as shown toÂ develop a constant potential gradient along the wire Two cells of emfs E _{1} and E_{2} are connected in series as shown in combinations (1) and (2). The balance points are obtained respectively at 400 cm and 240 cm from the point A. Find**

**(i)E**

_{1}/E_{2}.**(ii)balancing length for the cell E1 only.**

**Ans.**

**32.A number of identical cells, n each ofÂ emf E, Â internal Â resistance r,Â connected in series are charged by a DC source of emf E’, using a resistor, R.**

** (i)Draw the circuit arrangement.**

** (ii)Deduce the expressions for**

** (a)the changing current and**

** (b)the potential difference across the combination of the cells.Â [Delhi 2008]**

**Ans.**

**33.A potentiometer wire of length lm is connected to a driver cell of emf 3 V as shown in the figure. When a cell of 1.5 V emf is used in the secondary circuit, the balance point is found to be 60 cm. On replacing this cell and using a cell of unknown emf, the balance point shifts to 80 cm.**

** **

** (i)Calculate unknown emf of the cell.**

** (ii)Explain with reason, whether the circuit works, if the driver cell is replaced with a cell of emf 1 V.**

** (iii)Does the high resistance R, used in the secondary circuit affect the balance point? Justify your answer.[Delhi 2008]**

**Ans.**

**Ans**.**Â **(i) Net emf applied in the circuit = Applied potential difference – Total emf of all cells = 100 V-4 x 8 V = 68 V

**NOTE** During changing of the current, positive terminal of the battery is connected to positive terminal of the series combination of the cells.

**5 Marks Questions**

**35.(i) State the working principle of aÂ potentiometer. With the help of the circuit diagram, explain how aÂ potentiometer is used to compare the emf’s of two primary cells. Obtain the required expression used for comparing the emfs.**

** (ii) Write two possible causes for one sided deflection in a potentiometer experiment.[Delhi 2013]**

**Ans**.(i) **Working principle of potentiometer**Â When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

(ii) (a) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared.

(b) The positive ends of all cells are not connected to the same end of the wire

**36.State the working principle of aÂ potentiometer. Draw a circuit diagram to compare emf of two primary cells. Derive the formula used.**

** (i)Which material is used for potentiometer wire and why?**

** (ii)How can the sensitivity of a potentiometer be increased?Â [Delhi 2011]**

**Ans**.(i) **For working principle of potentiometer**

We use a long wire to have a lower value of potential gradient (i.e a lower “least count” or greater sensitivity of the potentiometer.

For circuit diagram to compare emf of two cells

(ii) Constantan or manganin (alloy) as they have low temperature coefficient of resistance.

(iii) The sensitivity of potentiometer can be increased by increasing the number of wires of potentiometer and hence, decreasing the value of potential gradient

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