CBSE Class 12 Physics Chapter-3 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 3 – Current Electricity 3 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-3 Important Questions
CBSE Class 12 Physics Important Questions Chapter 3 – Current Electricity
3 Marks Questions
1. What happens to the resistance of the wire when its length is increased to twice its original length?
Ans. 
Now 
1 = 2 and radius becomes r1
Since volume of the wire remains the same
New Resistance
= 
New Resistance becomes four times.
2. Mark the direction of current in the circuit as per kirchoff’s first rule. What is the value of main current in the shown network?
Ans. 
are in series
= 3+3=6
are in parallel
Net Current
I=
I= 
=1A
3.(a) Why do we prefer potentiometer to measure the emf of cell than a voltmeter?
(b) With suitable circuit diagram, show how emfs of 2 cells can be compared using a potentiometer?
Ans. (a) Since potentiometer is based on null method i.e. it draws no current from the
cell therefore potentiometer is preferred to measure the emf of a cell than a
voltmeter because emf of a cell is equal to terminal potential difference when no
current flows from the cell.
(b) Potentiometer works on the principle that when a constant current flows
through the wire of Uniform area of cross- section then
(Condition – close the switch and 3 such that 
comes in the circuit)
P.D. across AJ is 
Since no current flows between and 
= = 
= 
=
—-(1)
Close the switch 2 and 3 , cell 
comes in the circuit and balance point is obtained of 
= Since no current flows because A and are at same potential then 1 = E2
=
= = 
—-(2)
Comparing eg. (1) and (2)
4. Potential difference V is applied across the ends of copper wire of length (l) and diameter D. What is the effect on drift velocity of electrons if
(1) V is doubled
(2) l is doubled
(3) D is doubled
Ans.
(1) Since 
= 
V is doubled, drift velocity gets doubled.
(2) If l is doubled, drift velocity gets halved.
(3) Since V of is independent of D, drift velocity remains unchanged.
5. What is drift velocity? Derive expression for drift velocity of electrons in a good conductor in terms of relaxation time of electrons?
Ans. If is defined as the average velocity with which free electrons gets drifted in a direction opposite to that of electric field
If m is the mass of the electron and e be the charge of electron
Then on application of the electric field E, acceleration acquired by the electron is
first eg of motion 
since average initial velocity
u = O V= 
d
t = 
( relaxation time )
=>vd = a
wheree is the change on electron
E os the electric field intensity
is the relaxation time
m is the mass of electron.
6. The potentiometer circuit shown, the balance (null) point is at X. State with reason, where the balance point will be shifted when
(1) Resistance R is increased, keeping all parameters unchanged.
(2) Resistance S is increased, keeping R constant.
(3) Cell P is replaced by another cell whose emf is lower than that of cell Q.
Ans. (a) When resistance R is increased, the current through potentiometer wire AB will decrease, hence potential difference across A will decrease, so balance point shifts towards B.
(b) When resistance S is increased terminal potential difference of the battery will decrease, so balance point will be obtained at smaller length and hence shifts towards A.
(c) When cell P is replaced by another cell whose emf is lower than that of cell Q, the P.D. across AB will be less than that of emfQ so balance point will not be obtained.
7. (a) Using the principle of wheat stone bridge describe the method to determine the specific resistance of a wire in the laboratory. Draw the circuit diagram and write the formula used ?
In a whetstone bridge experiment, a student by mistake, connects key (k) in place of galvanometer and galvanometer (G) in place of Key (K). What will be the change in the deflection of the bridge.
Ans. (a) Close the Key (k) and jockey is moved along the wire till a certain point B is reached where galvanometer shows no deflection. Then the bridge is said to be balanced.
If Rcm is the resistance per can length of the wire then.
Since 
Where 
is the length of the wire.
(b) When the bridge is balanced, there will be no current in key, therefore constant current flows through the galvanometer and hence no change in deflection on pressing the key.
8. Two primary cells of emf’s 
are connected to the potentiometer wire AB as shown in the figure if the balancing length for the two combinations of the cells are 250 cm and 400 cm. find the ratio of.
Ans. 
—-(1)
—-(2)
Adding eg. (1) &(2)
= 250K+ 400K
= 250K + 400
= 650K
= 
= 325 K —–(3)
Subtracting eg. (1) & (2)
= 75K
=> 
9. Explain with the help of a circuit diagram, how the value of an unknown resistance can be determined using a wheat stone bridge?
Ans. Here P , Q , R are known resistance and X is an unknown resistance. Applying Kirchhoff’s law for closed path ABDA .
—–(1)
For closed path BCDB
—–(2)
Now the bridge is said to be balanced when
no current flows through the galvanometer
Ig = 0
Eg. (1) & (2) becomes
—–(3)
—–(4)
Equating (3) & (4)
10. Find the current drawn from a cell of emf IV and internal resistance 2/3 connected to the network shown in the figure. E = 1v r = 2/3
Ans. 
=>
=> R = 
=>R = 1
Now 1 , R and 1 are in parallel
=> 
I = 
I=1A
11. (a) State and explain kirchoff’s law?
(b) In the network shown, find the values of current
.
Ans. (a) Kirchoff’s first law – it states that the algebraic sum of the currents meeting
at a point in an electrical circuit is always zero.
Kirhoff’s second law – it states that in any closed part of an electrical circuit, the algebraic sum of emf & is equal to the algebraic sum of the products of resistances and current flowing through them for eg. For closed path ABCA
Or 
(b) Applying kirchoff’s law at point -D
For closed path ABDA
—–(2)
For closed path DBCD
Or 
—–(3)
Solving eg. (1) , (2) & (3)
I1 = 
12. The variation of resistance of a metallic conductor with temperature is given in figure.
(a) Calculate the temperature coefficient of resistance from the graph.
(b) State why the resistance of the conductor increases with the rise in temperature.
Ans. (a) Temperature coefficient of Resistance
Where R is the resistance of the conductor and is the temperature corresponding to pt.A
(b) Since R = 
When temperature increases, no of collisions increases average relaxation time decreases, hence resistance Increases.
13. A circle ring having negligible resistance is used to connect four resistors of resistances 6R , 6R , 6R and R as shown in the figure. Find the equivalent resistance. between points A & B
Ans.
6R , 6R and 6R are in parallel
Rs = 
Rs =2R => 2R and R are in series
14. A battery of emf E and internal resistance r sends a current 
, when connected to an external resistance of 
respectively. Find the emf. and internal resistance of the battery?
Ans. I1 = 
Similarly 
—(2)
From (1) & (2)
Emf. 
15. Find the value of unknown resistance X in the circuit shown in the figure if no current flows through the section AO. Also calculate the current drawn by the circuit from the battery of emf. 6v and negligible internal resistance.
Ans. As no current flows through AO then the circuit is said to be balanced wheat Stone bridge.
X = 
X = 6
Since in branch AO , I=0
Resistance of 10 between A and O is ineffective and the circuit reduce to
2
6 and 9 are in parallel
= 
and 2.4 are in parallel
Reff = 2.4 + 
Reff = 
Current I = 
=> I=1A
16. (a) Obtain ohm’s law from the expression for electrical conductivity.
(b) A cylindrical wire is stretched to increase its length by 10% calculate the
percentage increase in resistance?
Ans. (a) We know I = neAvd
J = 
Vol = 
=> J = 
Since J = 
E 
Let l and A be the length and area of the write.
I = JA
I = 
=> I = 
=> V = 
=> R = 
(specific resistance of a wire)
(b) 
Since volume of the wire remains the same
Al = 
Since R = 
Percentage increase in Resistance is
17. The current I flows through a wire of radius r and the free electron drift with a velocity 
what is the drift velocity of electrons through a wire of same material but having double the radius, when a current of 2I flows through it?
Ans. I = ne A vd
vd = 
= 
(1)
If vd’ is the drift velocity of electrons in the second wire
Vd’ = 
—–(2)
From eq . (1) & (2)
18. Three identical cells, each of emf. 2v and unknown internal resistance are connected in parallel .This combination is connected to a 5ohm resister. If the terminal voltage across the cell is 1.5volt. What is the internal resistance of each cell .hence define internal resistance of a cell?
Ans. E = 2v, V=1.5v, R = 5
Total internal resistance = 
Since r = 
r = 
The resistance offered by the electrolyte of the cell, when the electric current flows through it , is called as internal resistance of a cell.
19. Using Kirchhoff’s law, determine the current 
for the network shown.
Ans. Applying junction rule at point F
—-(1)
Loop rule for BAFCB
b —-(2)
Loop rule for FCDEF
—-(3)
solving eg . (1) , (2) & (3) we get
20. Show that when a current is divided between two resistances in accordance with kirchoff’s laws, the heat provided is minimum?
Ans. Consider two resistance 
in parallel and 
be the current. Using kirchoff’s first law
—-(1)
kirchoff’s second law
Heat produced in the circuit in t second is
(using eg.(1))
If the heat produced is minimum then 
This is in accordance with kirchoff’s law.
21. (a) Define emf. of a cell? On what factors does it depend?
(b) Figure below shows a 2.0v potentiometer used for the determination of internal resistance of a 1.5v cell. The balance point of the cell in open circuit is 76.3cm. When a resistance of 9.5
is used in external circuit of the cell the balance point shifts to 64.8cm length of the potentiometer. Determine the internal resistance of the cell.
Ans. (a) It is defined as the potential difference between the two electrodes of the cell in open Circuit (when no current is drawn) It depends on the following factors
(i) Nature of Electrodes
(ii) Nature and concentration of the Electrolytes
(iii) Temperature of the cell.
(b) Internal resistance of the cell.
r = R 
Here 
R = 9.5
=> r = 9.5 
22. A battery of emf 10 V and internal resistance 3 is connected resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Ans.Emf of the battery, E=10V
Internal resistance of the battery ,r =3
Current in the circuit, I=0.5A
Resistance of the resistor
The relation for current using Ohm’s law is,
Terminal voltage of the resistor=V
According to Ohm’s law,
V=IR
Therefore, the resistance of the resistor is 17 and the terminal voltage is 8.5v.
23. (a) Three resistors 
are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Ans.(a) Three resistors of resistances 
are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.
Total resistance=
(b) Current flowing through the circuit=I
Emf of the battery, E=12V
Total resistance of the circuit, R=
The relation for current using Ohm’s law is,
Potential drop across 
resistor = 
For Ohm’s law, the value of can be obtained as
Potential drop across 
resistor= 
Again, from Ohm’s law, the value of can be obtained as
Potential drop across 
resistor= 
Again, from Ohm’s law,the value of can be obtained as
Therefore, the potential drop across resistors are 2V, 4V, and 6V respectively.
24. At room temperature 
the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 
Ans.Room temperature, T=27
Resistance of the heating element at 
Temperature co-efficient of the material of the filament,
is given by the relation,
Therefore, at 1027
the resistance of the element is 117.
25. A negligibly small current is passed through a wire of length 15 m and uniform cross section 
,and its resistance is measured to be 5.0 . What is the material at the temperature of the experiment?
Ans. Resistivity of material can be calculated as:
Length of the wire, l=15m
Area of cross-section of the wire, a
Resistance of the material of the wire, R
Resistivity of the material of the wire=p
Resistance is related with the resistivity as:
Therefore, the resistivity of the material is 
26. A silver wire has a resistance of 2.1 Ω at
, and a resistance of 2.7 Ω at
. Determine the temperature coefficient of resistivity of silver.
Ans.Temperature, 
Resistance of the silver wire at 
Temperature, 
Resistance of the silver wire at 
Temperature coefficient if silver=
It is related with temperature and resistance as
Therefore, the temperature coefficient of silver is 0.0039
27. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 
Ans. Supply voltage, V=230V
Initial current drawn=
Initial resistance=
which is given by the relation,
Steady state value of the current, 
Resistance at the steady state=
which is given as
Temperature co-efficient of nichrome,
Initial temperature of nichrome, 
Study state temperature reached by nichrome= 
can be obtained by the relation for 
Therefore, the steady temperature of the heating element is 
28. A storage battery of emf 8.0 V and internal resistance 0.5 is being charged by a 120 V dc supply using a series resistor of 15.5. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Ans. Emf of the storage battery, E
Internal resistance of the battery, r=
DC supply voltage, V = 120 V
Resistance of the resistor, 
Effective voltage in the circuit =
R is connected to the storage battery in series. Hence, it can be written as
Current flowing in the circuit can be calculated as:
Voltage across resistor R is given by the product , IR= 7 
15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R
Terminal voltage of battery = 120-108.5 = 11.5 VA series resistor in a charging circuit limits the current drawn from the external source. The current will be
extremely high in its absence. This is very dangerous.
29. The number density of free electrons in a copper conductor estimated in Example is 
. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross section of the wire is 
and it is carrying a current of 3.0 A.
Ans. Number density of free electrons in a copper conductor, n=
Length of the copper wire, l=
Area of cross-section of the wire, 
Current carried by the wire, I=
, Which is given by the relation,
Where,
Electric charge=
Drift velocity
Therefore, the time taken by an electron to drift form one end of wire to the other is 
30. The earth’s surface has a negative surface charge density of 
. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different part of the globe [Radius of earth =
])
Ans. Surface charge density of the earth,
Current over the entire globe, 
Radius of the earth, r
Surface area of the earth,
Charge on the earth surface,
Time taken to neutralize the earth’s surface = tCurrent,
Therefore, the time taken to neutralize the earth’s surface is 282.77s.
31. Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 
Ans. (a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of
.
32. Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig 3.32.Each resistor has resistance.
Ans. The resistance of each resistor connected in the given circuit 
Equivalent resistance of the given circuit
The network is infinite, Hence, equivalent resistance is given by the relation,
Negative value of 
cannot be accepted. Hence, equivalent resistance,
Internal resistance of circuit is
Hence, total resistance of the given circuit 
Supply Voltage,
According to Ohm; s Law, current drawn from the source is given by the ratio,
33. Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 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 
?
Ans. Resistance of the standard resistor, R = 10.0
Balance point for this resistance, 
Current in the potentiometer wire = i
Hence, potential drop across R, 
Resistance of the unknown resistor = X
Balance point for this resistor, 
Hence, potential drop across X, 
The relation connecting emf and balance point is,
X=
=
Therefore, the value of the unknown resistance, X, is 11.75 .
If we fail to find a balance point with the given cell of emf, , then the potential drop across Rand X must be reduced by putting a resistance in series with it. Only if the potential drop across Ror X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.
34. Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
Ans.Internal resistance of the cell=
Balance point of the cell in open circuit,
An external resistance 
is connected to the circuit with 
New balance point of the circuit,
Current flowing through the circuit=
The relation connecting resistance and emf is,
Therefore, the internal resistance of the cell is
.