XII - SCIENCE PHYSICS SONAM TUITIONS

9. CURRENT ELECTRICITY
1. ROTATIONAL DYNAMICS
Q. Choose the correct option : (1 mark each)
(1) In a pokentiometer experiment, a cell balances against a length of 240 cm. When the cell in
shunted by 2, the balancing length becomes 120 cm. The internal resistance of the cell is
(July 22)
(a) 4  (b) 2  (c) 1  (d) 0.5 .
Ans.: (b) 2 
VERY SHOR ANSWER QUESTIONS
1. What is a potentiometer?
Ans.: A potentiometer is an instrument for measuring, comparing or dividing small potential
differences. It consists of a long and uniform resistance wire along which a potential gradient
is set up by connecting a cell of extremely stable emf connected across its ends.
2. Define potential gradient along a wire. (March 22)
Ans.: Potential gradient along a (potentiometer) wire is the potential difference (the fall of potential
from the high potential end) per unit length of the wire.
3. On what factors does the potential gradient of the wire depend?
Ans.: The potential gradient depends upon the potential difference between the ends of the wire
and the length of the wire.
4. What will be the effect on the position of null point on a potentiometer wire if the current
through the wire is decreased?
Ans.: The potential gradient along a potentiometer wire is directly proportional to the current through
the wire and the null length on a potentiometer is inversely proportional to the potential
gradient. Hence, the potential gradient decreases with a decrease in the current.
Consequently, the null length will decrease.
5. Why is no power consumed from the circuit containing unknown emf when a potentiometer
is used to measure the unknown emf?
Ans.: The adjustment of a potentiometer is a null comparison method. At balance, the unknown emf
and the potential difference across the balancing length of the potentiometer wire are in
opposition. Hence, the branch containing the unknown emf draws no current from the circuit at
the null point.
LAWS AND THEORIES
1. State Kirchhoff’s laws for electrical circuit. (July 22) ( 2 marks )
State the sign conventions used in these laws. (1½ marks)
Are the laws applicable to both AC and DC networks ? (½ marks)
Ans.: Kirchhoff’s circuital laws when steady currents are passing in an electrical network are as
follows :
Kirchhoff’s first law (or current law or junction law) : The algebraic sum of the currents at
any junction is zero.
n
 i  0
i 1

where  i is the current in the ith conductor meeting at the junction.

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XII - SCIENCE PHYSICS SONAM TUITIONS

Sign convention : A current entering the junction is taken as positive while a current leaving
the junction is taken as negative.
Kirchhoff’s second law (or voltage law or loop law) : Around a closed loop of an electrical
network, the algebraic sum of the emfs and the potential differences across all the circuit
elements in that loop is zero.
E   R  0
Sign convention : (a) In going round a loop, if we traverse a resistor in the direction of the
current through it, the potential difference ( p.d. ) across the resistor is taken as negative. The
p.d. is taken as positive if the direction in which we traverse the resistance is opposite to the
current through it. (b) The emf of a cell is taken as positive when we traverse the cell from its
negative terminal to the positive terminal while it is taken as negative when traversed in the
opposite direction.
Kirchhoff’s laws are applicable to both AC and DC circuits (networks). For AC circuits,
instantaneous currents and voltages are taken in the summations.

DERIVATIONS
1. Obtain the balancing condition in Wheatstone’s network. (3 marks)
2. Explain how a moving-coil galvanometer can be converted into a voltmeter. Derive the
necessary formula. (2 marks)
DISTINGUISH BETWEEN
1. Potentiometer and voltmeter. (March 22)
Ans.:
Potentiometer Voltmeter
1. A potentiometer is used to determine the 1. A voltmeter can be used to measure
emf of a cell, potential difference and the potential difference and terminal
internal resistance. voltage of a cell. But it cannot be used
to measure the emf of a cell.

2. Its accuracy and sensitivity are very high. 2. Its accuracy and sensitivity are less as
compared to a potentiometer.

3. It is not a portable instrument. 3. It is a portable instrument.

4. It does not give a direct reading. 4. It gives a direct reading.

SHORT ANSWER QUESTIONS

1. What are the disadvantages of a potentiometer (over a voltmeter)? (2 marks)
Ans.: Disadvantages of a potentiometer over a voltmeter :
(1) The use of a potentiometer is an indirect measurement method while a voltmeter is a
direct reading instrument.
(2) A potentiometer is unwieldy while a voltmeter is portable.
(3) Unlike a voltmeter, the use of a potentiometer in measuring an unknown emf requires a
standard source of emf and calibration.

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XII - SCIENCE PHYSICS SONAM TUITIONS

2. State the function of the shunt in modifying a galvanometer to an ammeter.

(½ mark each)
Ans.: Functions of the shunt in an ammeter :
(1) It lowers the effective resistance of the ammeter.
(2) It shunts off a larger fraction of the line current, thus protecting the sensitive meter
movement of the basic galvanometer.
(3) With a shunt of proper value, a galvanometer can be modified into an ammeter of
practically any desired range.
3. State the functions of the series resistance in modifying a galvanometer into a voltmeter.
(½ mark each)
Ans.: Functions of the high series resistance in a voltmeter :
(1) It increases the effective resistance of the voltmeter.
(2) It drops off a larger fraction of the measured potential difference, thus protecting the
sensitive meter movement of the basic galvanometer.
(3) With a resistance of proper value, a galvanometer can be modified to a voltmeter of any
desired range.

EXPERIMENTS & DIAGRAMS

1. Describe the construction of a metre bridge. (1 mark)
Explain with a neat circuit diagram, how you will determine an unknown resistance using a
metre bridge. (3 marks)
2. Describe Kelvin’s method to determine the resistance of a galvanometer by using a metre
bridge. (3 marks)
3. Describe the use of a potentiometer to compare the emfs of two cells by the direct method
(i.e., connecting them separately). (3 marks)
4. Describe the use of a potentiometer to compare the emfs of two cells by the sum and
difference method. (3 marks)
5. Describe with the help of a neat circuit diagram how you will determine the internal resistance
of a cell using a potentiometer. Derive the necessary formula. (3 marks)

INSTRUMENTS : CONSTRUCTION & WORKING

1. Explain the basic construction of a pivoted-type moving-coil galvanometer with a neat labelled
diagram. (2 marks)

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