EXERCISE
MULTIPLE CHOICE QUESTIONS
Choose the best possible answer:
1. Three resistors of resistance R each are combined in various ways. Which of the following cannot be obtained?
a. 3RQ b.(2R)/A * Omega
C. R/3 *D d. (2R)/3 * n
2. When n resistances each of value r are connected in parallel, then resultant resistance is x. When these n resistances are connected in series, total resistance is;
a. nx b. mx
c. x/n d. n'x
3. Two bulbs marked 200 watt-250 volts and 100 watt-250 volts are joined in series to 250 volts supply. Power consumed in circuit is;
a. 33 watt c. 100 watt
b. 67 watt d. 300 watt
4. Resistance of a wire is r ohms. The wire is stretched to double its length then its resistance in ohms is
a. r/2 b. 4r
c. 2r d. r/4
5. 10 electrons are moving through a wire per second, the current developed is
a. 1.6 * 10 ^ 19 * A b. 1 A
c. 1.6 x 10-13 A d. 106A
6. When a wire is stretched and its radius becomes r/2, then its resistance will be
a. 16 R b. 4R
d. 0 c. 2 R
7. How much voltage is required to make 2A flow through a resistance of 80?
a. 16 A Q¹ b. 160 A c. 16 AQ
d. [: A ^ - 1 * Omega ]
8. A wire of uniform cross-section A, length L and resistance R is cut into two equal parts. The resistivity of each part will be
a. doubled
b. halved
c. remains the same
d one fourth
9.
The resistivity of two wires is p; and p2 which are connected in series. If there dimensions are same then the equivalent resistivity of the combina
tion will be
a. (p: + P2) b. 1,1 P₁ P₂
C. P¹+p²|2 d. Pi P₂
10. The powers of two electric bulbs are 100 W and 200 W. Which are con nected to power supply of 220 V. The ratio of resistance of their filament will be,
a. 1:2 b. 2:1
c. 1:3 d. 4:3
11. Thermocouple is an arrangement of two different metals,
a. To convert heat energy into electrical energy
b. To produce more heat
c. To convert heat energy into chemical energy
d. To convert electric energy into heat energy
12. In order to achieve high accuracy, the slide wire of a potentiometer should be,
a. As long as possible
b. As short as possible
c. Very thin
d. Very thick
13. When a potentiometer is used for measurement of voltage of an unknown source, the power consumed in the circuit of the unknown source under null condition is,
a. high
b. small
c. ideally zero
d. very high
14. In a Wheatstone bridge method, the bridge is said to be balanced, when the current through the galvanometer is,
a. 1 A c. Maximum
b. 0 A d. Half of the maximum value
CONCEPTUAL QUESTIONS
Give a short response to the following questions:
1.A heavy duty battery of a truck maintains a current of 3A for 24 hour. How much charge flows from the battery during this time?
Ans. The amount of charge flows from the battery during 24 hours will be 259200 C
Calculation: Since it is given that the battery maintains a current of 3A for a time
of 24 hours, so to measure the amount of charge, we use the given equation.
AQ
At
⇒ 4Q=14t
By putting the values, we get;
AQ 3Ax24 hours
AQ-3Ax86400s 259200 As
24 hours 24x60x60's=86400s
=AQ-259200 C Thus the amount of charge that will flow from the truck battery during 24 hours, if it maintains a current of 3A will be 259200 C.
2 While analyzing a circuit the internal resistance of emf sources is ignored why? Ans. In a circuit analysis, the internal resistance of emf sources is usually ignored because its value is very small as compared to the load resistance.
Explanation: Every source of emf has an internal resistance r. No emf source with
out any internal resistance has been found. We know that an emf source provides voltage to the circuit. All the resistors in the circuit have a potential drop so that the sum of the individual potential drops is equal to the voltage of the emf source provided to the circuit. The internal re sistance behaves as it is connected in series with the source therefore; it also has a potential drop. This obviously affects the voltage strength of the battery, L.e. the voltage provided to the circuit is less than the voltage produced by the emf source. However, this effect (of voltage drop across the internal resistance) is so small that it is practically insignificant. The reason for less voltage drop is that the internal resistance of the emf source is always very small. For this reason, the resistance of the emf source is neglected in circuit analysis...
3. If aluminum and copper wires of the same length have the same resistance, which has the larger diameter? Why?
Ans. For the same length and same resistance, aluminum will has larger diameter.
Example: As we know that resistance of a conductor is directly proportional to its length and inversely proportional to the area of the conductor and is given by;
Where p is the resistivity of the material. Since aluminum has greater resistivity as compared to copper, therefore to make their resistances equal for the same length, area of aluminum will be taken larger. Since area of cross section is given by;
Therefore, diameter of aluminum will be larger as compared to copper.
4. Under what circumstances can the terminal P.D. of a battery exceed its emf?
Ans. When a battery is charging, then in that case the terminal P.D of the battery exceeds its emf.
Explanation: Terminal Potential Difference can be calculated by; VE-Ir (1)
Therefore, terminal P.D. can be small, equal or more than the emf of the source. When the circuit is open (no load) the emf and terminal voltage are same because there is no current in the circuit.
When there is load resistance in the circuit, terminal voltage is less than the emf by a factor Ir as shown in Eq. (1). However, terminal voltage will exceed the emf of the battery when current is flows in the reverse direction in the battery, therefore, the potential drop is posi tive and the terminal P.D. exceeds the emf by an amount Ir.driven backward through the battery; when it enters the positive terminal and leaves at negative terminal. This is possible when we put the battery to charge. In such a situation, the terminal potential is larger than the emf. Since the current
5. What is the difference between an emf and P.D.?
6. How the loop rule and junction rule are based on the conservation of energy principle?
Ans. Kirchhoff's Loop Rule (KVL):
Statement: In a closed electrical circuit, the algebraic sum of all the electromotive forces and the voltage drops in the resistors of the circuit is zero. Explanation: Consider the circuit shown in the figure. Let the current I is clockwise,
then by applying KVL, we get;
E-Ir-IR=0 E=Ir+IR (1)
Here & is the electromotive force in the circuit, is the current in the circuit, R is the load resistance and r is the internal resistance of the emf.
Kirchhoff's Loop Rule and Law of Conservation of Energy: Consider Eq. (1). The left side of the equation is the energy gained by a unit charge as it goes through the battery. The right side gives an account of the utilization of that energy; part of the energy gained (Ir), is dissipated into the battery and the rest, IR, in the external circuit. Thus the emf supplies energy to the charge as source of energy and the potential drop across various elements dissipate this en
ergy into other forms of energy.
Kirchhoff's Junction Rule (KCL):
Statement: The algebraic sum of all the currents flowing to ward a junction is equal to the sum of all the currents flow
ing away from the junction. Mathematical Form:
Σ1=0 = Σ-0 = ΣQ=0 →(2)
Kirchhoff's Junction Rule and Law of Conservation of Charge: Eq. (2) indicates thelaw of conservation of charge. The net charge at the junction is zero. Charge nelther accumulates nor produced at the junction. Therefore, charge is conserved.
7. Why rise in temperature of a conductor is accompanied by a rise in the resistance?
Ans. With the increase in temperature, the lattice atoms of a conductor start bration vigorously and thus the chance of collision of charge particles with these viatoms increases. This results in the increase of resistance.
Explanation: Resistance is the opposition offered by the conductor to the flow of
electrons. This resistance comes due to collision of conduction electrons with the
lattice atoms of the conductor and the bound electrons. Clearly, it depends upon the space among the lattice atoms on which the free electrons move. Now when the conductor heats up through any means, the atoms in the conductor gains energy and vibrate more vigorously with greater ampli tudes. This narrows the space for the free electrons to move and collide with the atoms more frequently as compared to the colder state. Therefore, the resistance. of the conductor increases.
8. Does the direction of emf provided by a battery depend on the direction of current flow through the battery?
Ans. No, direction of emf provided by the battery does not depend upon the di rection of flow of current through the battery but it depends on the internal composition of the battery.
Explanation: The direction of emf is decided by the construction of the battery cell (internal composition), it does not depend upon the direction of current flow. As emf source is responsible for the flow of current in a circuit, that's why we can say that, the direction of current flow in a circuit depends upon the direction of emf
does not depend upon the direction of current.
9. Voltages are always measured between two points. Why?
Ans. Since voltage is the measure of potential difference between two points, therefore, we take two points to measure voltages.
Explanation: The other name of voltage is the potential difference and it is the work done in moving a unit positive charge in moving from one point to the other point against electric field. It cannot be measured at a single point because it is a relative term. Even for calculating potential at a single point, we take reference point at infinity. Hence voltages are always measured between two points, not at a single point.
10. Is every emf a potential difference? Is every potential difference an emf?
Ans. Yes, every emf is a potential difference but not every potential difference is an emf.
Explanation: Since emf of a source can be defined as the work done in moving aunit charge from low potential to high potential on behalf of some internal energy stored in the source. Therefore, emf can be regarded as the potential difference between the two terminals of the source. This is not true in case of potential difference, actually potential difference between two points in the circuit is the work done by a charge in moving from high potential to a low potential by losing energy. Therefore, we cannot say that potential difference is equivalent to emf.
11. How much charge flow in pocket calculator each minute when the current is 0.0001 A?
Ans. The amount of charge flows in pocket calculator each minute will be 0.006 C.
Calculation: Since it is given that the current <= 0 for 1 minute, l.e. 60 degrees * s 50 to measure the amount of charge, we use the given equation.
I = (Delta*Q)/(Delta*t)
Rightarrow Delta*Q = I*Delta*t
By putting the values, we get; triangle Q = 0.0001A * 1min
Lambda * 0 = 0.0001A * 605 = 0.006As
= triangle Omega = 0.006C
Thus the amount of charge that will flows in the pocket calculator each minute
* 1min = 60s
when the current is 0.0001 A is 0.006 C.
12. When Wheatstone bridge is balanced, then no current flows through the galvanometer, why?
Ans.Under balanced condition, the potential difference across the galvanometer is zero; therefore no current flows through it and shows no deflection. Explanation: The circuit diagram of Wheatstone bridge is shown in the figure. Four resistance P, Q R and X are arranged in such a way that they form a closed loop ABCD. A single emf source & is attached with the circuit across point A and C, through a key. The galvanometer G is connected as a bridge between point B and D. The resistance of P and Q are fixed, whereas R can be varied, to determine the unknown resistance X. When the key closed the current is divided between the branch AB and AD as I and I_{2} respectively and galvanometer G will show deflection. The value of variable resistor R is adjusted until the galvanome ter G shows zero deflection. Under such condition the bridge is said to be balance. The galvanometer reads ZERO because point B and D are at same po tential. This means that voltage drops across AB and AD is equal.
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