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 5 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
5 Marks Questions
1. (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Ans. (a) There are three resistors of resistances,
They are connected in parallel. Hence, total resistance(R) of the combination is given by,
Therefore, total resistance of the combination is
.
(b) Emf of the battery, V=20V
Current 
flowing through resistor 
is given by,
Current 
flowing through resistor 
is given by,
Current 
flowing through resistor 
is given by,
Total current, 
Therefore, the current through each resister is 10A, 5A, and 4A respectively and the total current is 19A.
2. Determine the current in each branch of the network shown in fig 3.30:
Ans. Current flowing through various branches of the circuit is represented in the given figure.
Current flowing through the outer circuit
Current flowing through branch 
Current flowing through branc 
Current flowing through branch 
Current flowing through branch 
Current flowing through branch 
For the closed circuit ABDA, potential is zero i.e.,
For the closed circuit BCDB, potential is zero i.e.,
For the closed circuit ABCFEA, potential is zero i.e.,
From equations (1) and (2), we obtain
Putting equation (4) in equation (1), we obtain
It is evident from the given figure that,
Putting equation (6) in equation (1), we obtain
Putting equations (4) and (5) in equation (7), we obtain
Equation (4) reduces to
Therefore, current in branch
In branch
In branch
In branch
In branch
Therefore total current = 
3. (a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end when the resistor Y is of 12.5 Ω.
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Ans. A metre bridge with resistors X and Y is represented in the given figure.
(a) Balance point form end 
Resistance of the resistor 
Condition for the balance is given as,
Therefore, the resistance of resistor X is 
The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.
If X and Y are interchanged, 
get interchanged.
The balance point of the bridge will be 
from A.
= 100 − 39.5 = 60.5 cm
Therefore, the balance point is 60.5 cm from A.
When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.
4. (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 
are joined in series to provide a supply to a resistance of 8.5. What are the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
Ans. (a) Number of secondary cells, n
Emf of each secondary cell, E
Internal resistance of each cell, r= 0.015
series resistor is connected to the combination of cells.
Resistance of the resistor, R = 8.5
Current drawn from the supply=I, which is given by the relation,
Terminal voltage, V = IR = 1.39 
8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
After a long use, emf of the secondary cell, E=1.9V
Internal resistance of the cell, r= 380 Ω
Maximum current 
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.
5. Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (
Al = 
, Cu =
, Relative density of Al = 2.7, of Cu = 8.9.)
Ans.Resistivity of aluminium, 
Relative density of aluminium, 
Let 
be the length of aluminium wire and 
be its mass.
Resistance of the aluminium wire =
Area of cross-section of the aluminium wire = 
Resistivity of copper, 
Relative density of copper, 
Let 
be the length of copper wire and 
be its mass.
Resistance of the copper wire =
Area of cross-section of the copper wire = 
The two relations can be written as
……(1)
……(2)
It is given that,
And,
=
Mass of the aluminium wire,
……(3)
Mass of the copper wire,
……(4)
Dividing equation (3) by equation (4), we obtain
For 
For 
It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than copper.
Since aluminium is lighter, it is preferred for overhead power cables over copper.
6. Answer the following questions:
[a ] A steady current flows in a metallic conductor of non-uniform cross section of non-uniform cross section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
[b] Is Ohm’s law universally applicable for all conducting elements?
If not, give examples of elements which do not obey Ohm’s law.
[c] A low voltage supply from which one needs high currents must have very low resistance. Why?
[d] A high tension (HT) supply of, say, 6 kV must have a very large internal resistance
Ans.
[a] When a steady current flows in a metallic conductor of non-uniform cross section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.
[b]No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.
[c] According to Ohm’s law, the relation for the potential is V=IR
Voltage (V) is directly proportional to current (I).
R is the internal resistance of the source,
If V is low, then R must be very low, so that high current can be drawn from the source.
[d] In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.
7. [a] Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
[b] Given the resistances of 
how will be combine them to get an equivalent resistance of
[c] Determine the equivalent resistance of networks shown in fig.3.31.
Ans. (a) Total number of resistors= n
Resistance of each resistor
(i) When 
resistors are connected in series, effective resistance is the maximum, given by the product 
.
Hence, maximum resistance of the combination,
(ii) When n resistors are connected in parallel, the effective resistance 
is the minimum, given by the ration 
.
Hence, minimum resistance of the combination,
(iii) The ration of the maximum to the minimum resistance is,
The resistance of the given resistors is,
Equivalent resistance, 
Consider the following combination of the resistors.
Equivalent resistance of the circuit is given by,
Equivalent resistance, 
Equivalent resistance of the circuit is given by,
Equivalent resistance, 
Consider the series combination of the resistors, as shown in the given circuit.
Equivalent resistance of the circuit is given by the sum,
Consider the series combination of the resistors, as shown in the given ciruit. Equivalent resistance of the circuit is given by,
(c) (i) It can be observed form the given circuit that in the first small loop, two resistors of resistance 
each are connected in series.
Hence, their equivalent resistance=
It can also be observed that two resistors of resistance 
each are connected in series.
Hence, their equivalent resistance
Therefore, the circuit can be redrawn as
It can be observed that and 
resistors are connected in parallel in all the four loops.
Hence, equivalent resistance 
of each loop is given by,
The circuit reduces to
All the four resistors are connected in series.
Hence, equivalent resistance of the given circuit is 
(iii) It can be observed from the given circuit that five resistors of resistance 
each are connected in series.
Hence, equivalent resistance of the circuit 
8. Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40
maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replace by a cell of unknown emf and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
[a] What is the value 
?
[b] What purpose does the high resistance of 600 
have?
[c] Is the balance point affected by this high resistance?
[d] Is the balance point affected by the internal resistance of the driver cell?
[e] Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
[f] Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
Ans. [a] Constant emf of the given standard cell, 
Balance point on the wire, 
A cell of unknown emf,, replaced the standard cell. Therefore, new balance on the wire, 
The relation connecting emf and balance point is,
The value of unknown emf is 
[b] The purpose of using the high resistance of 600 is to reduce the current through the galvanometer when the movable contact is far from the balance point.
[c] The balance point is not affected by the presence of high resistance.
[d] The point is not affected by the internal resistance of the driver cell.
[e] The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.
[f] The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.
Modification: The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.