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1 Exercises Question 3.1: The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery? Answer 3.1: Emf of the battery, E = 12 V Internal resistance of the battery, r = 0.4 Ω Maximum current drawn from the battery = I According to Ohm s law, The maximum current drawn from the given battery is 30 A. Question 3.2: 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 3.2: Emf of the battery, E = 10 V Internal resistance of the battery, r = 3 Ω Current in the circuit, I = 0.5 A Resistance of the resistor = R The relation for current using Ohm s law is, Terminal voltage of the resistor = V According to Ohm s law, V = IR = = 8.5 V Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is 8.5 V. Question 3.3: a) Three resistors 1 Ω, 2 Ω, and 3 Ω 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. 1

2 Answer 3.3: (a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. Total resistance = = 6 Ω (b) Current flowing through the circuit = I Emf of the battery, E = 12 V Total resistance of the circuit, R = 6 Ω The relation for current using Ohm s law is, Potential drop across 1 Ω resistor = V 1 From Ohm s law, the value of V 1 can be obtained as V 1 = 2 1= 2 V (i) Potential drop across 2 Ω resistor = V 2 Again, from Ohm s law, the value of V 2 can be obtained as V 2 = 2 2= 4 V (ii) Potential drop across 3 Ω resistor = V 3 Again, from Ohm s law, the value of V 3 can be obtained as V 3 = 2 3= 6 V (iii) Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively. Question 3.4: (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. Answer 3.4: (a) There are three resistors of resistances, R 1 = 2 Ω, R 2 = 4 Ω, and R 3 = 5 Ω They are connected in parallel. Hence, total resistance (R) of the combination is given by, Therefore, total resistance of the combination is Ω. (b) Emf of the battery, V = 20 V Current (I 1) flowing through resistor R 1 is given by, Current (I 2) flowing through resistor R 2 is given by, Current (I 3) flowing through resistor R 3 is given by, 2

3 Total current, I = I 1 + I 2 + I 3 = = 19 A Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A. Question 3.5: At room temperature (27.0 C) 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 C 1. Answer 3.5: Room temperature, T = 27 C Resistance of the heating element at T, R = 100 Ω Let T 1 is the increased temperature of the filament. Resistance of the heating element at T 1, R 1 = 117 Ω Temperature co-efficient of the material of the filament, Therefore, at 1027 C, the resistance of the element is 117Ω. Question 3.6: A negligibly small current is passed through a wire of length 15 m and uniform crosssection m 2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment? Answer 3.6: Length of the wire, l =15 m Area of cross-section of the wire, a = m 2, Resistance of the material of the wire, R = 5.0 Ω Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as Therefore, the resistivity of the material is Ω m. 3

4 Question 3.7: A silver wire has a resistance of 2.1 Ω at 27.5 C, and a resistance of 2.7 Ω at 100 C. Determine the temperature coefficient of resistivity of silver. Answer 3.7: Temperature, T 1 = 27.5 C Resistance of the silver wire at T 1, R 1 = 2.1 Ω Temperature, T 2 = 100 C Resistance of the silver wire at T 2, R 2 = 2.7 Ω Temperature coefficient of silver = α It is related with temperature and resistance as Therefore, the temperature coefficient of silver is C 1. Question 3.8: 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 to a 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 C 1. Answer 3.8: Supply voltage, V = 230 V Initial current drawn, I 1 = 3.2 A Initial resistance = R 1, which is given by the relation, Steady state value of the current, I 2 = 2.8 A Resistance at the steady state = R 2, which is given as Temperature co-efficient of nichrome, α = C 1 Initial temperature of nichrome, T 1= 27.0 C Study state temperature reached by nichrome = T 2 T 2 can be obtained by the relation for α, Therefore, the steady temperature of the heating element is C 4

5 Question 3.9: Determine the current in each branch of the network shown in figure: Answer 3.9: Current flowing through various branches of the circuit is represented in the given figure. I 1 = Current flowing through the outer circuit I 2 = Current flowing through branch AB I 3 = Current flowing through branch AD I 2 I 4 = Current flowing through branch BC I 3 + I 4 = Current flowing through branch CD I 4 = Current flowing through branch BD For the closed circuit ABDA, potential is zero i.e., 10I 2 + 5I 4 5I 3 = 0 2I 2 + I 4 I 3 = 0 I 3 = 2I 2 + I 4 (1) For the closed circuit BCDB, potential is zero i.e., 5(I 2 I 4) 10(I 3 + I 4) 5I 4 = 0 5I 2 + 5I 4 10I 3 10I 4 5I 4 = 0 5I 2 10I 3 20I 4 = 0 I 2 = 2I 3 + 4I 4 (2) For the closed circuit ABCFEA, potential is zero i.e., (I 1) + 10(I 2) + 5(I 2 I 4) = 0 10 = 15I I 1 5I 4 3I 2 + 2I 1 I 4 = 2 (3) From equations (1) and (2), we obtain I 3 = 2(2I 3 + 4I 4) + I 4 I 3 = 4I 3 + 8I 4 + I 4 3I 3 = 9I 4 3I 4 = + I 3 (4) Putting equation (4) in equation (1), we obtain I 3 = 2I 2 + I 4 4I 4 = 2I 2 I 2 = 2I 4 (5) It is evident from the given figure that, I 1 = I 3 + I 2 (6) 5

6 Putting equation (6) in equation (1), we obtain 3I 2 +2(I 3 + I 2) I 4 = 2 5I 2 + 2I 3 I 4 = 2 (7) Putting equations (4) and (5) in equation (7), we obtain 5( 2 I 4) + 2( 3 I 4) I 4 = 2 10I 4 6I 4 I 4 = 2 17I 4 = 2 I 4 = 2 Equation (4) reduces to I 3 = 3I 4 = 3 ( 2 17 ) = 6 I 2 = 2I 4 = 2 ( 2 17 ) = 4 I 2 I 4 = 4 17 ( 2 17 ) = 6 I 3 + I 4 = ( 2 17 ) = 4 I 1 = I 3 + I 2 = = 10 Therefore, the current in branch in AB = I 2 = 4 The current in branch in BC = I 2 I 4 = 6 In branch CD = I 3 + I 4 = 4 In branch AD = I 3 = 6 In branch BD = I 4 = 2 Total current = I 1 = 10 Question 3.10: (a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is 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? (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? Answer 3.10: A metre bridge with resistors X and Y is represented in the given figure. (a) Balance point from end A, l 1 = 39.5 cm Resistance of the resistor Y = 12.5 Ω Condition for the balance is given as, Therefore, the resistance of resistor X is 8.2 Ω. 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. 6

7 (b) If X and Y are interchanged, then l 1 and 100 l 1 get interchanged. The balance point of the bridge will be 100 l 1 from A. 100 l 1 = = 60.5 cm Therefore, the balance point is 60.5 cm from A. (c) 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. Question 3.11: 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 3.11: Emf of the storage battery, E = 8.0 V Internal resistance of the battery, r = 0.5 Ω DC supply voltage, V = 120 V Resistance of the resistor, R = 15.5 Ω Effective voltage in the circuit = V 1 R is connected to the storage battery in series. Hence, it can be written as V 1 = V E V 1 = = 112 V Current flowing in the circuit = I, which is given by the relation, Voltage across resistor R given by the product, IR = = V DC supply voltage = Terminal voltage of battery + Voltage drop across R Terminal voltage of battery = = 11.5 V A 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. Question 3.12: 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 3.12: Emf of the cell, E 1 = 1.25 V Balance point of the potentiometer, l 1= 35 cm. The cell is replaced by another cell of emf E 2. New balance point of the potentiometer, l 2 = 63 cm Therefore, emf of the second cell is 2.25V. 7

8 Question 3.13: The number density of free electrons in a copper conductor estimated in Example 3.1 is m 3. 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 m 2 and it is carrying a current of 3.0 A. Answer 3.13: Number density of free electrons in a copper conductor, n = m 3 Length of the copper wire, l = 3.0 m Area of cross-section of the wire, A = m 2 Current carried by the wire, I = 3.0 A, which is given by the relation, I = naev d Where, e = Electric charge = C V d = Drift velocity Therefore, the time taken by an electron to drift from one end of the wire to the other is s. 8

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General Physics (PHY 2140) Lecture 9 Electrodynamics Electric current temperature variation of resistance electrical energy and power http://www.physics.wayne.edu/~apetrov/phy2140/ Chapter 17-18 1 Department

### Direct Current Circuits. February 18, 2014 Physics for Scientists & Engineers 2, Chapter 26 1

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### 1) Two lightbulbs, one rated 30 W at 120 V and another rated 40 W at 120 V, are arranged in two different circuits.

1) Two lightbulbs, one rated 30 W at 120 V and another rated 40 W at 120 V, are arranged in two different circuits. a. The two bulbs are first connected in parallel to a 120 V source. i. Determine the

### Electric currents (primarily, in metals)

Electric currents (primarily, in metals) Benjamin Franklin was experimenting electricity in the mid- XVIII Century. Nobody knew if it was the positive charges or negative charges carrying the current through

### 2. Q. 1 to 5 are Very short Answer type questions (1 Mark each. ) 3. Q. 6 to 12 are short Answer type questions. (2 Marks each. )

1. All questions are compulsory. 2. Q. 1 to 5 are Very short Answer type questions (1 Mark each. ) 3. Q. 6 to 12 are short Answer type questions. (2 Marks each. ) 4. Q. 13 to 24 are short answer questions

### 1 Fig. 3.1 shows the variation of the magnetic flux linkage with time t for a small generator. magnetic. flux linkage / Wb-turns 1.

1 Fig. 3.1 shows the variation of the magnetic flux linkage with time t for a small generator. 2 magnetic 1 flux linkage / 0 10 2 Wb-turns 1 2 5 10 15 t / 10 3 s Fig. 3.1 The generator has a flat coil

### 2/25/2014. Circuits. Properties of a Current. Conservation of Current. Definition of a Current A. I A > I B > I C B. I B > I A C. I C D. I A E.

Circuits Topics: Current Conservation of current Batteries Resistance and resistivity Simple circuits 0.1 Electromotive Force and Current Conventional current is the hypothetical flow of positive charges

### AP Physics C - E & M

AP Physics C - E & M Current and Circuits 2017-07-12 www.njctl.org Electric Current Resistance and Resistivity Electromotive Force (EMF) Energy and Power Resistors in Series and in Parallel Kirchoff's

### To receive full credit, you must show all your work (including steps taken, calculations, and formulas used).

Page 1 Score Problem 1: (35 pts) Problem 2: (25 pts) Problem 3: (25 pts) Problem 4: (25 pts) Problem 5: (15 pts) TOTAL: (125 pts) To receive full credit, you must show all your work (including steps taken,

Electromagnetic Fields Lecture 5 The Steady Current Field What is current? Electric current: Flow of electric charge. Electric current in metals A solid conductive metal contains free electrons. When a

### Circuits. PHY2054: Chapter 18 1

Circuits PHY2054: Chapter 18 1 What You Already Know Microscopic nature of current Drift speed and current Ohm s law Resistivity Calculating resistance from resistivity Power in electric circuits PHY2054:

### Review of Ohm's Law: The potential drop across a resistor is given by Ohm's Law: V= IR where I is the current and R is the resistance.

DC Circuits Objectives The objectives of this lab are: 1) to construct an Ohmmeter (a device that measures resistance) using our knowledge of Ohm's Law. 2) to determine an unknown resistance using our

### Chapter 17. Current and Resistance. Sections: 1, 3, 4, 6, 7, 9

Chapter 17 Current and Resistance Sections: 1, 3, 4, 6, 7, 9 Equations: 2 2 1 e r q q F = k 2 e o r Q k q F E = = I R V = A L R ρ = )] ( 1 [ o o T T + = α ρ ρ V I V t Q P = = R V R I P 2 2 ) ( = = C Q

### PEP 2017 Assignment 12

of the filament?.16.. Aductile metal wire has resistance. What will be the resistance of this wire in terms of if it is stretched to three times its original length, assuming that the density and resistivity

### Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

PCE CURRENT ELECTRICITY C Electric Current Current is rate of flow of charge from one region to another. Mathematically, dq I where dq is net dt q charge transported at section during time dt. For steady

### Revision Compare Between. Application

evision Compare etween Points of Comparison Series Connection Parallel Connection Drawing otal resistance ( ) = + + 3 3 Potential Difference () = + + 3 = = = 3 Electric Current (I) I = I = I = I 3 I =

### Electric current is a flow of electrons in a conductor. The SI unit of electric current is ampere.

C h a p t e r at G l a n c e 4. Electric Current : Electric current is a flow of electrons in a conductor. The SI unit of electric current is ampere. Current = Charge time i.e, I = Q t The SI unit of charge

### Electric Currents and Circuits

Nicholas J. Giordano www.cengage.com/physics/giordano Chapter 19 Electric Currents and Circuits Marilyn Akins, PhD Broome Community College Electric Circuits The motion of charges leads to the idea of

### Electric Current & DC Circuits

Electric Current & DC Circuits Circuits Click on the topic to go to that section Conductors Resistivity and Resistance Circuit Diagrams Measurement EMF & Terminal Voltage Kirchhoff's Rules Capacitors*

### Current and Resistance

Current and Resistance 1 Define the current. Understand the microscopic description of current. Discuss the rat at which the power transfer to a device in an electric current. 2 2-1 Electric current 2-2

### coil of the circuit. [8+8]

Code No: R05310202 Set No. 1 III B.Tech I Semester Regular Examinations, November 2008 ELECTRICAL MEASUREMENTS (Electrical & Electronic Engineering) Time: 3 hours Max Marks: 80 Answer any FIVE Questions