Bihar Board - Class 12 Physics - Chapter 3: Current Electricity Long Answer Question
Long Question Answer
1. Explain Ohm’s Law.
Answer:
This Ohm’s Law states that the current that flows through a conductor is directly proportional to its potential difference applied across its ends, here the temperature and physical conditions remain unchanged. Here the current is directly proportional to voltage through a resistor, which means the current doubles then the voltage also doubles. To make current flow there should be a voltage across that resistance. Ohm’s Law states the relationship between these terms: voltage V, current I and resistance R. This can be written as:
V ∝ I
Or I ∝ V
V= IR
Where R resistance is constant. The value of it depends upon the nature, cross-section, length and temperature conductor. Here, V = potential difference is in Volt, I= current is in Ampere and R= resistance (constant). The SI unit of resistance is Ohm and denoted by Omega (Ω). The necessary condition of Ohm’s Law for applications like
To find out the voltage, resistance or current of an electric circuit.
It is also used to find the desired voltage drop across the electric components.
It is also used in devices like dc ammeter and other dc shunts to divert the current.
For Ohm’s Law, it is important that the temperature should be constant. It is right only for electric elements. For elements like capacitance, resistance, etc. the voltage and current will not be constant with respect to time. Ohm’s Law is also not applicable for unilateral elements like diodes and transistors because here current flows in one direction only.
2. Explain Wheatstone Bridge.
Answer:
The Wheatstone Bridge is an arrangement made up of four resistances connected with each other to form the arms of shape like a quadrilateral A battery with key and galvanometer are connected at its two diagonal positions. Where R1 andR3 are standard arms or ratio arms, R2 is known resistance and R4 is unknown resistance. The bridge is balanced when
Vb=Vd
Here, there is no flow of current through the galvanometer.
The Wheatstone bridge is known to be sensitive because there will be a slight change of resistance in the galvanometer due to ample deflection. For the sensitivity of any galvanometer, the magnitude of all four resistances should be in the same order as A, B, C, and D.
3. 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?
Answer: Potentiometer :- A potentiometer is a device used to measure potential difference.
Principle. When a current flows through a wire of uniform thickness, the potential difference between its two points is directly proportional to the length of the wire between these two points.
V∝ l
V =Kl
… when [K is called construction potential gradient]
Two factors :
(a) Potential gradient
(b) Length of potentiometer wire.
(i) When R is increased, the balance point will shift towards B.
(ii) If resistance S increased keeping ‘R’ constant, the balance points will not change.
4. Explain Kirchhoff's current law.
Answer: There are two Kirchhoff’s laws and they are Kirchhoff’s current law and Kirchhoff’s voltage law.
Kirchhoff’s current law:- Kirchhoff’s junction rule is also said to be Kirchhoff’s point rule, it is the algebraic sum of currents of conductors which are in a network and meeting at a point is zero. Basically, the sum of currents entering the junction is equal to the sum of leaving currents at that junction.
As per figure A, the sum will be i1 + i2 = i3 .
As per figure B, the sum will be i1 = i2 + i3+ i4
As per figure C, the sum will be i1 + i2 + i3 = 0.
In the last given figure, we can observe that the current flows in and doesn’t go out. Here the directions for every current problem are solved arbitrarily. So here the problem is solved, some currents have negative value, which shows that the actual current will flow in the opposite direction as compared with arbitrarily chosen initially. If the current value is positive then the direction of the current is the same as chosen initially.
5. Explain Kirchhoff's voltage law.
Answer: Kirchhoff's voltage law:- Kirchhoff’s voltage law is also said to be Kirchhoff's loop rule. Here the law states that the sum of electromotive forces in a loop is equal to the sum of potential drops in the loop. It can also be said as the directed sum of voltages is zero around every closed loop. It can be known from the below-given figure.
Here the potential difference can be written as, Vb - Va = E1. As per the figure.
As for the potential difference Vc - Vd is written as -E2. i,e; Vc - Vd = -E2. Using Ohm’s Law, Vb - Vc = iR1, and Vd - Va = iR2. These four relations of equations make loop equations as E1 - E2 - iR1 - iR2 = 0. Here, R1 and R2 are values of resistances in ohms and EMF (electromotive forces) as E1 and E2 in volts, the current value I is obtained here. If E2 > E1 then the current will be negative and it indicates the current flows in the opposite direction. This law can be understood from the following figure as well.
6. Define resistivity of a conductor. Plot a graph showing the variation of resistivity with temperature for a metallic conductor. How does one explain such a behavior, using the mathematical expression of the resistivity of a material?
Answer:
(i) Resistivity of conductor : It is the resistance of a conductor of unit length and unit area of cross-section.
The S.I. unit of resistivity is m (ohm-meter)
= RAl
(ii) Variation of resistivity with temperature :
The resistivity of a material is given by
= mne2
On increasing temperature, the average speed of drifting electrons increases. As a result collisions are more frequent. Average relaxation time τ decreases, hence ‘ρ’ increases.
7. Find the value of current I1 which is flowing in the circuit given below. Find the value by using Kirchhoff’s rules.
Answer:
By Kirchhoff’s first law at junction E, we get current
I3 = I1 + I2
By Kirchhoff’s second law in the loop ABCDA
- 20 I2 + 30 I1 = 0
2 I2 – 3 I1= 8 [ by 10] ... (1)
In loop ABFEA, we get
80 - 20 I2 +20 - 20 I3 =0
I2 – I3 = 5 [ by 20] ... (2)
Putting the value of I3 into (2), we have
I2+ (I2 + I2 )=5 => 2 I2 + I1 =5 … (3)
Solving equations (2) and (3), we get
I1 = - 34 A
I1 = – 0.75 A
Here the current value obtained is negative means the direction is opposite as per the figure given above.
8. What are the differences between e.m.f. and terminal potential difference?
Answer: e.m.f.
(1) It is defined as the potential difference between the two terminals of a cell when it is in the open circuit.
(2) It is independent of the resistance of the external circuit.
(3) It is a cause of electric current.
(4) e.m.f. of a cell is greater than the potential difference between the two terminals of the cell.
(5) It is used as a source.
Terminal Potential difference :
(1) It is defined as the potential difference between the two terminals of a cell when it is in the closed circuit.
(2) Potential difference between any two points of a circuit is proportional to the resistance between these two points.
(3) It is an effect of electric current.
(4) It is less than the e.m.f. of the cell.
(5) It can be measured between any two points of the circuit.
9. (a) You are required to select a carbon resistor of resistance of 56 kΩ ± 10% from a shopkeeper. What would be the sequence of colour bands required to code the desired resistor?
(b) Write two characteristic properties of the material of a meter bridge wire.
(c) What precautions do you take to minimize the error in finding the unknown resistance of the given wire?
Answer:
(a) Number 5 corresponds to green, No. 6 corresponds to blue, 103 corresponds to orange and 10% corresponds to silver.
∴ Sequence of Colours is
Green, blue, orange, silver
(b) Metre bridge wire must have
High resistivity
Low-temperature coefficient of resistivity
(c) To minimize the error in determining the resistance of a wire, the Wire should be of uniform thickness The balance point should be near the midpoint of the wire.
10. Distinguish between resistance and specific resistance.
Answer:
Resistance | Specific resistance |
1. It is the ratio of the P.D. across the conductor to the two ends of a conductor to the current flowing i.e. R = VI | 1. It is the resistance of conductor of unit length and unit area of cross-section and is given by ρ = RAI |
2. It depends upon the length, area of cross-section and nature of material. | 2. It is independent of length and area of cross-section of the material and depends only on the nature of the material. |
3. Unit of resistance in SI is ohm (Ω). | 3. Unit of specific resistance in SI is ohm-meter (Ωm) |
11. 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?
Answer:
Emf of battery, E = 8 V
Internal resistance of battery, r = 0.5 Ω
Supply Voltage, V = 120 V
The resistance of the resistor, R = 15.5 Ω
Let V' be the effective voltage in the circuit.
Now, V' = V - E
V' = 120 - 8 = 112 V
Now, current flowing in the circuit is:
I = V' / (R + r)
I = 11215.5+0.5 = 7A
Now, using Ohm ’s Law:
Voltage across resistor R is v = IR
V = 7 x 15.5 = 108.5 V
Now, the voltage supplied, V = Terminal voltage of battery + V
The purpose of having a series resistor is to limit the current drawn from the supply.
12. Differentiate ohmic and non- ohmic conductors.
Answer: Differences between ohmic and non-ohmic :-
Ohmic conductors | Non-ohmic conductors |
1. They obey Ohm’s law. | 1. They do not obey Ohm’s law. |
2. The graph between V and I is a straight line passing through the origin. | 2. Graph between V and I is not a straight line. |
3. The value of resistance is constant and is independent of V and I. | 3. The value of resistance is not constant and changes when voltage is applied. |
4. At constant temperature all metals are ohmic conductors. | 4. Vacuum tubes, semiconductors, transistors, electrolytes etc, are non-ohmic conductors. |
.
13. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 , what is the maximum current that can be drawn from the battery?
Answer :
Given, the emf of battery, E = 12 V
Internal resistance of battery, r = 0.4 Ohm
Let I be the maximum current drawn from the battery.
We know, according to Ohm's law
E = Ir
I = E/r = 12/0.4 =30 A
Hence the maximum current drawn from the battery is 30 A.
14. A battery of emf 10 V and internal resistance 3 Ω is connected to a 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?
Answer:
Given, The emf of the battery, E = 10 V
The internal resistance of the battery, r = 3 Ohm
Current in the circuit, I = 0.5 A
Let R be the resistance of the resistor.
Therefore, according to Ohm's law:
E = IR' = I(R + r)
10 = 0.5(R + 3)
R = 1 Ohm
Also,
V = IR (Across the resistor)
= 0.5 x 17 = 8.5 V
Hence, terminal voltage across the resistor = 8.5 V
15. What do you understand about the internal resistance of a cell ? On what factors does It depend on and how ?
Or
What do you mean by internal resistance of a cell ? Write the factors affecting it.
Answer:
The resistance offered by the electrolyte of the cell during the flow of current inside the cell is called its internal resistance.
The following factors affect the internal resistance:
Distance between the electrodes:
As the distance increases, the internal resistance increases.
Area of the immersed electrodes:
As the area increases, the internal resistance decreases.
Concentration of the electrolyte:
As the concentration is more, the internal resistance is more.
Temperature:
The increase of temperature decreases the internal resistance.
16. If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Answer: Since the resistances are in series, the current through each one of them will be equal to the current through the circuit but voltage/ potential drop will be different.
Total resistance, R = 6 Ω
Emf, V = 12 V
According to Ohm's law:
V = IR
12 = I x 6
I = 12/6 = 2 A
Now, using the same relation, voltage through resistors:
1 Ω : V(1) = 2 x 1 = 2V
2 Ω : V(2) = 2 x 2 = 4V
3 Ω : V(3) = 2 x 3 = 6V
(Note: V(1) + V(2) + V(3) = 2 + 4 + 6 = 12 V )
17. In a meter bridge , the balance point is found to be at 39.5 cm from the end A, when the resistor Y determines the current in each branch of the network shown in Fig. 3.30:of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
Answer:
Balance point from the end A, l1= 39.5 cm
Resistance of Y = 12.5 Ω
We know, for a meter bridge, balance condition is:
XY = l1l2 = l1100-l1
X = 39.5100-39.5 12.5 =8.2
The connections between resistors in a Wheatstone or meter bridge made of thick copper strips to minimize the resistance of the connection which is not accounted for in the bridge formula.
18. What is Ohm’s law ? On what factors does the resistance of a conductor depend upon ?
Answer: If all the physical conditions of any conductor as length, temperature, etc. remain constant, then the current which flows through it is proportional to the potential difference applied across the ends of the conductor. If I is the current in conductor and V is the potential difference, then
V ∝ l = V=RI
Where, R is a constant, called resistance of conductor.
For factors affecting resistance of conductor:
Length: The resistance of conductor is directly proportional to the length of the conductor i.e., R ∝ l
Area of cross – section: The resistance of a conductor is inversely proportional to the area of cross – section of the conductor i.e., R ∝lA
Temperature: Increase in temperature, increases the resistanceCombining above two laws, we get,
R ∝lA
R = lA
Where, = a constant, called specific resistance of the material of the conductor.
Specific resistance:
We have,
R = lA
Let l = 1 And A = l, then
R = P
Thus, the specific resistance of a material is defined by the resistance of unit length and unit area of cross-section of that material.
Unit: Now,
= RAl = Ohm x m = Ω x m
19. What are the possible errors of a meter bridge and how can they be removed ?
Answer: The possible errors and their removal methods are:
1. It might happen that the wire is not uniform. To remove this error, balance points should be obtained at the-middle.
2. During The experiment, it is assumed that the resistance of L shaped plates are negligible, but actually it is not so. The error created due to this is called end error. To remove this error, the resistance box and the unknown resistance must be interchanged and then the mean reading should be taken.
3. If the jockey is pressed for a long period of time, then it gets heated and its resistance changes. Hence, the jockey must not be pressed for a long interval.
20. 109 electrons flow from point A to B in 10-3 second. Find out magnitude and direction of electric current.
Answer: Formula: I = net
Given: n = 109t = 10-3 sec.
Putting the given value in the formula, we get
I = 1091.610-1910-3
or I = 1.6 x 10-7 ampere.
Direction of electric current will be from B to A.
21. 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?
Answer:
E ∝ l
E1 E2 = l1l2
∴ E2 = l2l1 x E1 = 6335 x 12.5
i.e., E2 = 2.25 V
22. 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?
Answer:
r = 0.5 Ω, R = 15.5 Ω
when the storage battery emF 8 V is charged with a d.c supply of 120V the net EMF of the circuit
E= 120 – 8 = 112V
Therefore the current in the circuit during charging,
I = ER+r = 11215.4+0.5 = 7A
The terminal voltage of the storage battery would be equal to the sum of its EMF and the potential difference across its internal resistance i.e. terminal voltage = 8 + 0.5 x 7 = 11.5 V
23. (a) What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate Runknown by any other method?
(b) What is the advantage of using thick metallic strips to join wires in a potentiometer?
Answer: (a) In the case of a Wheatstone bridge, the technique of null point uses balanced Wheatstone bridges, in which the galvanometer’s resistance does not influence the balance point. There is no requirement to find the currents in the galvanometer and resistances. The unbalanced Wheatstone bridge can also be utilized to determine the unknown resistance. However, in this technique, we need the additional accurate measurement of every current in the galvanometer and resistors as well as the galvanometer’s internal resistance.
(b) Resistance is given by the relation
R = lA
So, as the cross-sectional area of the wire increases, the resistance of the wire reduces. Therefore, the metal strips possess very low resistance and can be easily neglected while determining the wire’s length used to find the null point.
24.(a) Three resistors IΩ, 2Ω, and 3Ω 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.
Answer:
(a) In series, R = R1 + R2 + R3 = 1 + 2 + 3 = 6Ω
(b) V = IR, I = VR = 126 = 2A
Potential drop across IΩ = V1 = IR1= 2 x 1 = 2 V
Potential drop across 2Ω = V2 = 2 x 2 = 4 V
Potential drop across 3Ω =V3 = 2 x 3 = 6V
25. Answer the following questions.
(a) A steady current flows in a metallic conductor of the 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 internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
Solution: (a) Except current the values of all the other quantities depend upon the area of cross-section of the conductor. Hence, only current remains constant, when it flows through a conductor of the non-uniform area of the cross section.
(b) No, ohm’s law is not obeyed by all the elements. For example, vacuum diode tube and semiconductor diodes.
(c) The maximum current that can be drawn from a voltage supply is given by,
Imax =Er
Obviously, Imax will be large, if r is small.
(d) If the circuit containing the H.T supply gets short-circuited accidently, the current in the circuit will not exceed the safe limit, in case the internal resistance of the H.T supply is very large.
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