Please refer to Assignments Class 10 Science Electricity Chapter 12 with solved questions and answers. We have provided Class 10 Science Assignments for all chapters on our website. These problems and solutions for Chapter 12 Electricity Class 10 Science have been prepared as per the latest syllabus and books issued for the current academic year. Learn these solved important questions to get more marks in your class tests and examinations.
Electricity Assignments Class 10 Science
Very Short Answer Type Questions :
Question. Name the device/instrument used to measure potential difference. How is it connected in an electric circuit?
Answer : The device which is used to measure potential difference is voltmeter. Voltmeter is connected in parallel in an electric circuit.
Question. How much current will an electric bulb of resistance 1100 W draw from a 220 V source? If a heater of resistance 100 W is connected to the same source instead of the bulb, calculate the current drawn by the heater.
Answer : Resistance of bulb, R = 1100 W
V = 220 volt
V = IR or I = V/R
I = 220/1100 = 1/5 A
When heater is connected with the same source then
I = V/R = 220/100 = 2.2 A
Question. State the factors on which the heat produced in a current carrying conductor depends. Give one practical application of this effect.
Answer : We know that H = VIt or H = I2Rt
Heat produced in a current carrying conductor
H ? I2 (Square of the current in the circuit)
? R (Resistance of the conductor)
? t (Time for which current is passed in conductor)
This effect applied in electric heater.
Question. Draw a schematic diagram of an electric circuit consisting of a battery of two cells each of 1.5 V, 5 W , 10 W and 15 W resistors and a plug key, all connected in series.
Answer : Schematic diagram of electric circuit con¬taining cells, key and three resistances.
Question. Given below is a circuit showing current flowing in it.
Identify each component A, B, C, D of this circuit. 38
Answer : Circuit
A — Bulb (load)
B — Rheostat
C — Cell
D — Ammeter (being in series)
Question. A large number of free electrons are present in metals yet no current flows in the absence of electric potential across it. Explain the statement with reason.
Answer : Though there are large number of free electron present in a conductor, but their motion is random motion in the absence of potential difference. Their average velocity is zero. Hence there is no current flowing in conductor. But when a potential difference is applied across the ends of the conductor, it sets the electrons to move in a direction. The motion of charge produces an electric current in the conductor.
Question. Out of the two wires X and Y shown below. Which one has greater resistance. Justify your answer.
Answer : If X and Y are of same material then length of wire X = l
Area of cross-section = A
Rx = P,l/A
Similarly, Ry = p,21/A
Ry/Rx = 2
or RY = 2RX
wire Y has two times resistance than that of wire X .
Question. How would the reading of voltmeter (V) change, if it is connected between B and C? Justify your answer.
Answer : 1W,3Ω,2Ω resistances are connected in series.
R = R1+ R2+ R3
= 1 + 3 + 2 = 6 Ω
I = V/R = 3/6 = 1/2A
Current in each resistance is same, i.e. 1/2A
Voltage across B & C = 1/2 x 3 = 3/2 Volt.
VBC = 3/2 Volt.
Question. State a difference between the wire used in the element of an electric heater and in a fuse wire.
Answer : The wire used in element of electric heater has high resistance and high melting point where as a fuse wire has a low resistance and low melting point.
Question. Power of a lamp is 60 W. Find the energy in joules consumed by it in 1 s.
Answer : P = 60W, t = 1 s
Energy = ^VI ht
E = P X t = 60×1 J
E = 60 J
Question. Elements of electric toasters and electric iron are made of an alloy rather than a pure metal. Give two reasons to justify the statement.
Answer :
a. Alloys have higher resistivity than their constituents pure metals.
b. At high temperature alloys do not oxidise.
Question. Out of the two, a toaster of 1 kW and an electric heater of 2 kW, which has a greater resistance?
Answer : As R V2/P . If V is constant. Then R ∝ 1/P i.e. more power, lesser be the resistance and vice versa.
Toaster has lesser power (1 KW) than electric heater (2 KW) therefore resistance of toaster is more than resistances of heater.
Question. The amount of charge passing through a cell in four second is 12 C. Find the current supplied by cell.
Answer : Given: t = 4 s
Q = 12C
I = Q/t = 12/4 A = 3 A
Question. List in a tabular form two differences between a voltmeter and an ammeter.
Answer :
Question. A thick wire and a thin wire of the same material are successively connected to the same circuit to find their respective resistance. Which one will have lower resistance? Give reason.
Answer : As resistance ∝ 1/A i.e. more area of cross-section lesser the resistance and vice versa. So thick wire has lower resistance.
Question. In the circuit diagram shown, the two resistance wires A and B are of same area of cross-section and same material, but A is longer than B. Which ammeter A1 or A2 will indicate higher reading for current? Give reason.
Answer : Length of A is greater than B. Area of cross-section of A and B is same.
Therefore resistance of A 2 resistance of B6a R ? l@.
Current in A is lesser than current in B. Hence A2 will give higher reading.
Question. Mention the condition under which charges can move in a conductor. Name the device which is used to maintain this condition in an electric circuit.
Answer : When a potential difference is applied across the two ends of the conductor, charges (electrons) will move in the conductor.
Potential difference is applied by a cell to maintain the charge to move.
Question. Three V-I graphs are drawn individually for two resistors and their series combination. Out of A, B, C which one represents the graph for series combination of the other two. Give reason
Answer : More slope of V-I graph means more resistance, slope of C is maximum. Hence its resistance is maximum.
So it is for series combination of two resistors.
Question. (a) What are the values of mA and m A?
(b) Draw the symbols of battery and rheostat.
Answer : 1 µ A = 1 milli ampere = 10-3 A
1 µ A = 1 micro ampere = 10-6 A
Question. Give reason for the following:
a. Tungsten used almost exclusively for filament of electric lamp.
b. Why do we use copper and aluminium wires for transmission of electric current?
Answer :
a. Tungsten is used in making the filament of electric lamp because it has high resistivity and high melting point.
b. The copper and aluminium have low resistivity and high conductivity.
Question. What is the resistance of an ideal voltmeter?
Answer : The resistance of an ideal voltmeter is infinite.
Question. How are ammeters and voltmeters connected in a circuit? What do they help us measure?
Answer : An ammeter which measure the current in a circuit is connected in series. Voltmeter is used to measure potential difference across a conductor so it is connected in parallel to it.
Question. The following table gives the resistivity of three samples:
Which of them is suitable for heating elements of electrical appliances and why?
Answer : The resistivity of C sample is maximum so it is suitable for making heating element.
Question. a. Give reason why tungsten is used for making filament of electric lamps.
b. The elements of heating electrical appliances are made-up of an alloy rather than pure metal.
Answer : a. Due to (1) high resistivity and high melting point tungsten is used for making filaments of electric lamp.
b. high resistivity, and high melting point than their constituents pure metals which do not oxidise at high temperature.
Question. Mention two special features of the material to be used as element of an electric iron.
Answer : A material to be used as element of an electric iron must have (i) high melting point, (ii) high resistivity.
Question. B1, B2 and B3 are three identical bulbs connected as shown in the figure. When all the three bulbs glow, a current of 3 A is recorded by the ammeter A.
a. What happens to the glow of the other two bulbs when the bulb B1 gets fused?
b. What happens to the reading of A1 A2, A3 and A when the bulb B1 gets fused?
Answer :
a. The other two bulbs will glow even if B1 gets fused. There is no change in glow of B2 and B3.
Here V = 4.5 V
and I = 3 A
RP = V/I = 4.5/3
RP = 1.5 W
For parallel combination
Question. (a) Draw a circuit diagram to show how two resistors are connected in series.
(b) In a circuit, if the two resistors of 5 ohm and 10 ohm are connected in series, how does the current passing through the two resistors compare?
Answer :
b. In series combination of resistances, the current remains same.
Question. A bulb is rated at 5.0 V, 100 mA. Calculate its (a) power and (b) resistance.
Answer : Rating of bulb, V = 5.0 Volt.
I = 100 mA
I = 100 X 10-3 A
I = 0.1A
a. Power of bulb = V X I
P = 5.0 X 0.1W = 0.5W
b. V = IR,
R = V/I = 5.0 /0.1 Ω = 50 Ω
Question. An electric bulb draws a current of 0.2 A when the voltage is 220 volts. Calculate the amount of charge flowing through it in one hour.
Answer : Given: I = 0.2 A
V = 220 Volt
t = 1 hr.
Q = ?
I = Q/t or Q = I x t
Q = 0.2A x 1 hr = 0.2x60x60 A – s = 720 C
Question. Find the resistance of bulb rated as 100W-250V.
Ans : We have, P = 100W
V = 250 Volt
P = V2/R or R = V2/P
R = (250)2/100 = 250 x 250/100 = 625 Ω
Question. Why do the wires connecting an electric heater to the mains not glow while its heating element does?
Answer : Connecting wire has low resistivity or good conductivity, i.e. resistance of these wires is negligible hence no heat is produced while heating element is of an alloy whose resistivity is high, due to high resistance heat is produced in the element.
Question. State the factors on which the resistance of a cylindrical conductor depends. How will resistance of a conductor change if it is stretched so that its length is doubled?
Answer : Resistance of cylindrical conductor depends upon its length and cross- sectional area.
When conductor is stretched its radius decreases but the volume of the conductor in both the cases will be same. If length is stretched to twice.
l2 = 2l1
As l1A1 = l2A2
l1A1 = 2l1A2
A1/A2 = 2
R1 = P,l1/A1
and R2 = P,l2/A2 = P,2h/A2
R2/R1 = 2 x A1/A2 = 2 x 2 = 4
R2 = 4R1
Short Answer Type Questions :
Question. Find the equivalent resistance of the following circuit:
Answer : In the given circuit 2 W,2 W resistances are in parallel.
Rp = 1Ω
1W, 1W resistances also are in parallel.
Rpl = 0.5 Ω
The circuit can be reduced as,
Now all resistances are in series combination.
R = 3 + 3 + 1 + 0.5 = 7.5 Ω
Question. Study the following circuit and answer the following questions:
a. State the type of combination of the two resistors in the circuit.
b. How much current would flow through: (1) 10 Ω resistor and (2) 15 Ω resistor?
c. What would be the ammeter reading?
Answer :
a. 10 Ω and 15 Ω are in parallel combination.
b. Potential difference across each is 3V.
I in 10 Ω resistor = 3/10 = 0.3A
I in 15 Ω resistor = 3/15 = 0.24
I = 0.3 + 0.2 = 0.5 A
Long Answer Type Questions :
Question. (a) For the circuit shown in the diagram, calculate:
(i) value of current through the 30 W resistor.
(ii) total resistance of the circuit.
(b) Give two advantages of connecting electrical devices in parallel with battery.
Answer :
b. (i) Voltage across each appliance remains
(ii) If any appliance fails to work other appliances continue to work.
Question. A 4 W resistance wire is doubled on itself. Calculate the new resistance of the wire.
Answer : L et the initial length = l
Initial resistance = 4 W
when it is doubled on itself, length becomes half and area becomes double (=2A)
Question. An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V.
What are the current and the resistance in each case?
Answer : As P = VXI
a. When heating is at maximum rate
P = 840W
Question. In a household, 5 tube lights of 40 W each are used for 5 hours and an electric press of 500 W for 4 hour everyday. Calculate the total energy consumed by the tube lights and press in a month of 30 days.
Answer :
Power of 1 tube = 40W
Power of 5 tubes = 5x40W = 200W
Energy consumed by 5 tubes in 5 hr. per day
= 200×5 = 1000Wh
Energy consumed by electric press per day
= 500Wx4 hr
= 2000Wh
Total energy consumed per day
= ^1000 + 2000hWh
= 3000Wh = 3 kWh
Total energy consumed in 30 days
= 3X30 kWh
= 90 kWh
Question. State Ohm’s law. Write the mathematical representation of Ohm’s law. Use this relationship to define 1 ohm. List two disadvantages of connecting different electrical appliances in series.
Answer :
If the physical conditions of a conductor are kept constant then current is directly proportional to the potential difference applied across it.
Mathematical representation of Ohm’s law V = IR.
V/I = R
1 Volt/1Ampere = 1 ohm
If by applying a potential difference of 1 volt across a conductor, the current is 1A then the resistance of the conductor is said to be 1 ohm.
Two disadvantages of connecting electrical appliance in series.
a. If one appliance fails to operate then the circuit is broken and other devices also will not operate.
b. Different devices require different amount of current to operate but in series combination, same current is supplied to all electrical appliances.
Question. A piece of wire is redrawn, without change in volume so that its radius is halved. Compare the new resistance with the original resistance.
Answer :
Question. The potential difference across the terminals of a cell is 1.5 volt. If it is connected with a resistance of 30 ohms, calculate the current flowing through the circuit.
Answer : Given: V = 1.5 volt.
R = 30 Ω
I = V/R = 1.5/30 = 1/20 = 0.05 A
Question. Two electric lamps of 100 W and 25 Ω respectively are joined in parallel to a supply of 200 V. Calculate the total current flowing through the circuit.
Answer : Bulbs 100 Ω 25Ω
V = 200 V 200 V
Total P = P1+ P2
= 100 + 25 = 125Ω
I = P/V = 125/200 A
I = 0.625 A
Question. Two identical resistors, each of resistance 2 Ω are connected in turn: (i) in series, and (ii) in parallel to a battery of 12 V. Calculate the ratio of power consumed in the two cases.
Answer : Given: R1 = R2 = 2 Ω
(i) In series RS = 2 + 2 = 4 Ω
Power consumed in series:
Question. In the given circuit, calculate (a) total resistance of the circuit, and (b) current shown by the ammeter.
Answer :
a. 3 Ω and 2 Ω resistors are in series their effective resistance 3 + 2 = 5 W is in parallel to 5 W resistance.
Net resistance = 5/2 = 2.5 Ω
b. I = V/R = 2.5/2.5
I =1A
Question. (a) Write two points of difference between electric energy and electric power. (6) Out of 60 W and 40 W lamps, which one has higher electrical resistance when in use.
(c) What is the commercial unit of electric energy? Convert it into joules.
Answer: (a) Difference between electric energy and electric power:
i.e. less the power of electrical device, higher is its electrical resistance.
(c) Kilowatt hour – Commercial unit of electrical energy
1 kWh = 1000 Wh = 1000 J/S x 3600 sec
= 3600000 J = 3.6 x106J
Question. State Ohm’s law. Write the necessary conditions for its validity. How is this law verified experimentally? What will be the nature of graph between potential difference and current for a conductor? Name the physical quantity that can be obtained from this graph.
Answer: Ohm’s law : When the physical conditions such as temperature etc. remain same, the current flowing through the conductor is directly proportional to the potential difference applied across the ends of the conductor, i.e.,
Necessary condition for validity of Ohm’s law is that physical condition such as temperature of the conductor remains same.
Procedure:
(i) Complete the circuit by connecting one cell in the gap XY. Plug the key.
(ii) Note the reading in the ammeter for the current I and in the voltmeter for the potential difference, (V) across the nichrome wire.
(v) Find the ratio of F to / for each observation.
(vi) Plot a graph between V (y-axis) and I (x-axis).
Observation :
1. Voltmeter and ammeter reading increases as the number > of cells increase in series.
2. Same value of V/I is obtained in each case.
3. V-I graph is a straight line passing through the origin of the graph as shown.
Conclusion : Straight line nature of graph shows that the current is proportional to the potential difference. Hence, Ohm’s law verified. The slope of V-I graph gives the value of resistance of the conductor at the given temperature.
Question. Draw a labelled circuit diagram showing three resistors R1, R2 and R3 connected in series with a battery (E), a rheostat (Rh), a plug key (K) and an ammeter (A) using standard circuit symbol. Use this circuit to show that the same current flows through every part of the circuit.
List two precautions you would observe while performing the experiment.
Answer:
Aim: Same current flows through every part of the above circuit.
Procedure:
1. Connect ammeters, ‘A’1 between B and C, and ‘A2’ between D and E.
2. Adjust the sliding contact of the rheostat initially for a small current.
3. Note all the ammeter readings. These reading give us current flowing through the resistors R1, R2 and R3
4. The current in the circuit is now increased by changing the position of sliding contact J’ of the rheostat.
5. Note all the ammeter readings each time.
Conclusion: Same reading of all the ammeter in each observation concluded that same current flows through every part of the circuit.
Precautions:
1. All the connection should be tight and properly connected as per circuit diagram.
2. The positive terminal of the ammeter and voltmeter must be connected to the positive terminal of the battery or battery eliminator.
Question. Two wires A and B are of equal length and have equal resistance. If the resistivity of A is more than that of B which wire is thicker and why? For the electric circuit given below calculate:
(i) Current in each resistor,
(ii) Total current drawn from the battery, and
(iii) Equivalent resistance of the Circuit
Answer:
So, for different materials having same resistance per unit length, greater resistivity material wire has more cross-sectional area.
Hence, wire A is thicker than that of B.
(i) Current through each resistor
Question. Study the I-V graph for four conductors A, B, C and D having resistance RA,RB, Rc and RD respectively, and answer the following questions:
(i) Which one of these is the best conductor?
(ii) f all the conductors are of same length and same material, which is the thickest?
(iii) If all the conductors are of same thickness and of same material, which is the longest?
(iv) If the dimensions of all the conductors are identical, but their materials are different which one would you use as (a) resistance wire (b) connecting wire?
(v) Which one of the following relations is true for these conductors?
(vi) If conductors A and B are connected in series and I-F graph is plotted for the combination, its slope would be
(a) less than that of A. (b) more than that of A.
(c) between A and B. (d) more than that of D.
(vii) If conductors C and D are connected in parallel and I-V graph is plotted for the combination, its slope would be
(a) lesser than that of A. (b) more than that of D.
(c) between C and D. (d) between B and C.
Answer:
(iii) Resistance oC length. So, A is the longest.
(iv) (a) Resistance wire – A (b) Connecting wire – D
(v) Option (a) is correct.
(vi) (a) In series, resistances are added. (Rs =RA + RB). So, in the given I-V graph, slope of series combination would be less than that of A.
(vii) (b) In parallel combination, the equivalent resistance is less than the least value resistance in the circuit. So, in the given I-V graph, slope for parallel combination is more than that
of D
Question. (a) Calculate the resistance of 1 km long copper wire of radius 1 mm. Resistivity of the copper is 1.72 x 10-8 Ω m.
(b) Draw a schematic diagram of a circuit consisting of a battery of 4 cells of 2V each connected to a key, an ammeter and two resistors of 2 Ω and
3 Ω respectively in series and a voltmeter to measure potential difference across 3
Answer:
Question. When a high resistance voltmeter is connected directly across a resister its reading is 2 V. An electric cell is sending the current of 0.4 A, (measured by an ammeter) in the electric circuit in which a rheostat is also connected to vary the current.
(a) Draw an equivalent labelled circuit for the given data.
(b) Find the resistance of the resister.
(c) Name and state the law applicable in the given case. A graph is drawn between a set of values of potential difference (V) across the resister and current (I) flowing through it. Show the nature of graph thus obtained.
Answer:
(c) Ohm’s law : When the physical conditions such as temperature etc. remain same, the current flowing through the conductor is directly proportional to the potential difference applied across the ends of the conductor, i.e.,
where R is constant of proportionality and is called resistance of the wire. Since current varies linearly with potential difference, the graph between V and I will be a linear in nature as shown
Question. Three bulbs each having power P are connected in series in an electric circuit. In another circuit, another set of three bulbs of same power are connected in parallel to the same source.
(i) Will the bulbs in both the circuits glow with the same brightness? Justify your answer.
(ii) Now let one bulb in each circuit get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
(iii) Representing each bulb by a resistor, draw circuit diagram for each case.
Answer: (i) Bulbs in parallel provide more illumination. This is because
(a) each bulb gets same voltage and is equal to the applied voltage.
(b) each bulb draws required current from the mains. Hence, they work properly.
(ii)When one bulb in each circuit get fused,
In series: Rest of the bulbs will not glow. This is because in series arrangement, there is only a single path for the flow of current.
In parallel: Rest of the bulbs will continue to glow as in parallel connection, (a) individual branch in the circuit completes its own circuit, or (b) different paths are available for the flow of current.
(iii) Circuit diagram
Question. (a) Though same current flows through the electric line wires and the filament of bulb, yet only the filament glows. Why?
(b) The temperature of the filament of bulb is 2700°C when it glows. Why does it not get burnt up at such high temperature?
(c) The filament of an electric lamp, which draws a current of 0.25 A is used for four hours. Calculate the amount of charge flowing through the circuit.
(d) An electric iron is rated 2 kW at 220 V. Calculate the capacity of the fuse that should be used for the electric iron.
Answer: (a) Electric line wires offer extremely low resistance to the flow of current, so they do not glow because negligible heat is produced in it.
The filament of bulb glows because it becomes red hot due to large amount of heat produced, as it offers high resistance to the flow of current through it.
(b) The filament of bulb when it glows at 2700°C does not get burnt because the tungsten metal of filament has
(i) a very high melting point (of 3380°C) and
(ii) a high resistivity. img
(c) Given: I = 0.25 A, t = 4 h = 4 x 60 x 60 sec.
So, amount of charge flowing the filament of electric lamp
So, the capacity of the fuse that should be used for the electric iron is of the order of 10 A.
Question. What is meant by electric current? Name and define its SI unit. In a conductor electrons are flowing from B to A. What is the direction of conventional current?
A steady current of 1 ampere flows through a conductor. Calculate the number of electrons that flows through any section of the conductor in 1 second. (Charge on electron 1.6 X 10-19 coulomb).
Answer: Electric Current: The amount of charge ‘Q’ flowing through a particular area of cross section in unit time ‘t’ is called electric current, i.e.
Electric current, I = Q/t
SI unit of electric current is ampere.
One ampere of current is that current which flow when one coulomb of electric charge flowing through a particular area of cross-section of the conductor in one second, i.e. 1A = 1 Cs-1. The direction of conventional current is A to B, i.e. opposite to the direction of flow of electrons. In a metal, flow of electrons carrying negative charge constitutes the current. Direction of flow of electrons gives the direction of electronic current by convention, the direction of flow of positive charge is taken as the direction of conventional current.
Charge = q = ne
Question. What is meant by electrical resistivity of a material? Derive its S.I. unit. Describe an experiment to study the factor on which the resistance of a conducting wire depends.
Answer: Mathematically, resistivity of the conducting material is given by
p = R x A/J
If l = 1 m, A = 1 m2, then p = R
Hence, the resistivity of the material is defined as the resistance offered by a metallic wire having a unit length and a unit area of cross-section. Since unit length and unit area of cross-section forms a cube, the specific resistance or resistivity can also be defined as the resistance offered by a cube of a material of side 1 m when current flows perpendicularly through the opposite faces. In SI system, its units is img
Experiment:
Aim : To study the factors on which resistance of conducting wires depends. Apparatus Required : A cell, an ammeter, nichrome wires of different length but same area of cross-section (thickness), nichrome wires of same length but different thickness, copper and iron wire of the same length and same thickness as that of any nichrome wire.
Procedure :
1. Connect the cell, an ammeter and plug key in series with nichrome wire of length T (marked 1) in the gap XY as shown.
2. Close the key and note the reading of ammeter. It measures the current ‘I’1 through the nichrome
wire (marked ‘1’).
3. Replace the marked 1 wire with another nichrome wire having same area of cross-section (thickness)
but of double length’2l’ (marked 2).
4. Note the ammeter reading (I2) again after closing the key.
5. Again replace the marked 2 wire with marked 3 wire which has the same length but is thicker than marked 1 and 2 nichrome wires. Again note down the current (I3) through this wire.
6. Unplug the key. Remove marked 3 nichrome wire from the gap XY. Connect the copper wire marked 4 having same length and same area of cross-section as that of nichrome wire marked 1. .
7. Plug the key again and note the ammeter reading. It measures the current (I4) through copper wire.
8. Repeat the experiment with iron wire and measure the current (I5).
Observation :
1. Current ‘ I’ is half ofI1 i.e., I2 =1/2 I1
2. Current I3 increases when thicker wire of same length and of same material
i. e., nichrome is used.
3. Current I4 and I5 is different for copper and iron wire.
Conclusion :
1. Different wires drew different amount of current from the same cell.
2. First observation indicates that the resistance of the conductor increases with increase in length. So, resistance is directly proportional to length.
3. Second observation shows that thicker wires have lesser resistance. So, resistance is inversely proportional to area of cross section of the wire.
4. Third observation shows that resistance of the conductor depends on the nature of its material.
Question. Derive the expression for the heat produced due to a current T flowing for a time interval ‘£’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is the relation known? How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V?
Answer: Heat produced in a conductor: Consider a wire AB having a resistance ‘R’ connected across the terminals of a cell. Let V be the potential difference applied by cell across the ends of a wire.
Let W be the work done in carrying the charge q across the conductor, then
Question. (a) Define electric power. Express it in terms of potential difference V and resistance R. (b) An electrical fuse is rated at 2 A. What is meant by this statement?
(c) An electric iron of 1 kW is operated at 220 V. Find which of the following fuses that respectively rated at 1 A,3 A and 5 A can be used in it.
Answer:
(a) Electric power: It is the rate of doing work by an energy source or the rate at which the electrical energy is dissipated or consumed per unit time in the electric circuit is called electric power.
(b) It means, the maximum current will flow through it is only 2 A. Fuse wire will melt if the current exceeds 2 A value through it. |
Question. Deduce the expression for the equivalent resistance of the parallel combination of three resistors R1, R2 and R3
Consider the following electric circuit:
(a) Which two resistors are connected in series?
(b) Which two resistors are connected in parallel?
(c) If every resistor of the circuit is of 2 Ω, what current will flow in the circuit?
Answer: Consider the following parallel circuit shown below: Let I1, I2 and I3 be the current flow through the resistor R1, R2 and R3 connected in parallel.
Using Ohm’s law, current through each resistor is
Question. (a) Two resistors R1and R2 may form (i) a series combination or (ii) a parallel combination, and the combination may be connected to a battery of 6 volts. In which combination, will the potential difference acrossR1 and across R2 be the same and in which combination, will the current through R1 and through R2 be the same?
(b) For the circuit shown in this diagram, calculate
(i) the resultant resistance.
(ii) the total current.
(iii) the voltage across 7 Ω resistor.
Answer: (a) Potential difference across R1 and R2 is same in parallel combination of R1 and R2 and the current through R1 and R2 will be same when they are connected in series.
Question.
In the given circuit, connect a nichrome wire of length ‘L’ between points X and Y and note the ammeter reading.
(i) When this experiment is repeated by inserting another nichrome wire of the same thickness but twice the length (2L), what changes are observed in the ammeter reading?
(ii) State the changes that are observed in the ammeter reading if we double the area of crosssection without changing the length in the above experiment. Justify your answer in both the cases.
(b) “Potential difference between points A and B in an electric field is 1 V”. Explain the above statement.
Answer: (a)(i) Ammeter reading in the second case is half of the ammeter reading in first case. This is because => i.e. when length is doubled, the resistance is doubled, this means the current is halved.
(ii) On doubling the area of cross-section without changing the length of the conductor, twice of the previous reading is observed in the ammeter. This is because So, when A is double, resistance becomes half, current will be doubled. (b) “Potential difference between points A and B in an electric field is 1 V”. It means 1 J work is done in moving 1 C of charge from point B to point A in an electric field.
Question. Explain with the help of a labelled circuit diagram, how will you find the resistance of a combination of three resistors, of resistance R1, R2 and R3 joined in parallel. Also mention how will you connect the ammeter and the voltmeter in the circuit while measuring the current in the circuit and the potential difference across one of the three resistors of the combination.
Answer: Parallel Combination:
1. Connect the three given resistor R1, R2 and R?i in parallel between the point XY with a battery, a plug key and ammeter in series as shown in figure.
2. Connect voltmeter in parallel with these resistors between the terminals X and Y.
3. Close the key and note the ammeter and voltmeter reading. Ammeter shows the total current drawn by the parallel combination of these resistors while voltmeter shows the voltage applied across the combination.
4. Using Ohm’s law, find the equivalent resistance of the combination, i.e. equivalent resistance,
5.
To find the current flow through any one of the resistor, ammeter will be connected in series with that resistor and to measure the potential difference across that resistor, voltmeter must be connected in parallel with that resistor as shown.
Question. An electric heater rated 600 W operates 6 hours per day. Find the cost to operate it for 30 days at Rs. 3.00 per unit.
Answer : Given: P = 600W
t = 6 hr per day
no. of days = 30
Energy consumed by the heater = PXt
= 600 W X 6,hr/day x 30 days
= 108000 Wh
= 108 kWh
Total cost @₹3/- = ₹108×3 = ₹324
Question. Three resistors of 5 W, 10 W, and 15 W are connected in series and the combination is connected to battery of 30 V. Ammeter and voltmeter are connected in the circuit. Draw a circuit diagram to connect all the devices in proper correct order. What is the current flowing and potential difference across 10 W resistance?
Answer : Effective resistance of the circuit
R = 5 + 10 + 15 = 30 Ω 139
Current in the circuit, I = V/R = 30/30 A
I = 1A
Current in 10 W resistor = 1A
Potential difference across 10 W resistor
V = IR = 1X10
V = 10 Volt
Question. Find the equivalent resistance across the two ends A and B of this circuit.
Answer :
The pairs of Rl and R2; R3 and R4 ; R5 and R6 and R7, and R8 are in parallel Their equivalent resistance = 1W each.
Now R12 and R34 are in series R14 = 1 + 1 = 2 W and R56 and R78 are in series
R56 = 1 + 1 = 2 Ω
Hence network can be reduced further as
Now R14 and R58 are in parallel
Net equivalent resistance of the network
= 1/2 + 1/2 = 1Ω
Question. A hot plate of an electric oven, connected to a 220 V line. It has two resistance coils A and B each of the 30 W which may be used separately, in series or in parallel. Find the value of the current required in each of the three cases.
Answer : V = 220 Volt
RA = RB = 30 W
a. When both are used separately.
Then current drawn by each
I = V/R = 220/30 = 7.3 A
b. When two coils are connected in series then total resistance of coils
= 30 + 30 = 60 Ω
I = 220/60 = 3.66 A
c. When two coils are connected in parallel then total resistance
= 30/2 = 15 A
Now current, I = 220/15 = 14.67 A
Question. A piece of wire having resistance R is cut into four equal parts,
a. How will the resistance of each part compare with the original resistance?
b. If the four parts are placed in parallel, how will the joint resistance compare with the resistance of the original wire?
Answer :
a. When wire is cut in equal pieces then resistance on one piece = π/4
b. Effective resistance in parallel combination of these four pieces of resistance π/4 each.
1/RP = 4/R + 4/R + 4/R + 4/R = 16/R
RP = R/16
Question. The rating of an electric heater is 1100 W, 220 V.
Calculate its resistance when it operates at 220 V.
Also, calculate the energy consumed in kWh in the month of November, if the heater is used daily for four hours at the rated voltage.
Answer :
P = 1100W
V = 220 V
R = V2/P = 220 X 220/1100 Ω = 484/11 Ω = 44 Ω
Total energy consumed in the month of Nov. (i.e. 30 days) @ 4 hr per day.
E = PXt
= 1100X4XhrX30
= 1100X120Wh
E = 132 kWh
Question. An electric heater rated 1200 W operates 2 hour per day. Find the cost of the energy required to operate it for 30 days at Rs 5.00 per unit.
Answer : Rating of heater = 1200W
t1 = 2 hr per day
Total time for 30 days
= t1x30 = 2X30
= 60 hrs
Energy consumed in one month
= PXt = 1200×60
= 72000 Wh
= 72 kWh
Cost of energy consumed in 30 days
= ₹72X5= ₹360
Question. Find out the following in the electric circuit given in figure:
a. Effective resistance of two 8 W resistors in the combination.
b. Current flowing through 4 W resistor.
c. Potential difference across 4 W resistor.
d. Power dissipated in 4 W resistor.
e. Difference in ammeter readings, if any.
Answer :
b. Total resistance of the circuit
= (4 + 4) W = 8 W
Current in 4 W resistor
I = 8/8 = 1A
c. Potential difference across 4 W resistor
V = IR = 1×4 = 4 Volt.
d. Power dissipated in 4 W resistor
P = I2R = (1)2X4W
P = 4W
e. Both the ammeters are in series in the circuit so current will be same in both ammeters. Hence no difference in the reading of two ammeter.
Question. Calculate the resistance of an electric bulb which allows a 10 A current when connected to a 220 V power source.
Answer : I = 10 A
V = 220 Volt.
Resistance of bulb = V/I = 220/10 = 22Ω
Question. An electric lamp is marked 100 W, 220 V. It is used for 5 hours daily. Calculate: (i) its resistance while glowing, (ii) energy consumed in kWh per day.
Answer :
Electric Iron 100W –– 220 V
Used for time, t = 5 hr. daily
(i) R = V2/P = 220 X 220/100 = 484 Ω
(ii) Energy consumed by iron per day
= PXt = 100X5 Wh
= 500Wh = 0.5 kWh
Question. A torch bulb is rated 5 V and 500 mA. Calculate its (i) power (ii) resistance and (iii) energy consumed when it is lighted for 4 hours.
Answer : Given: 5V––500 mA
(i) Power = V#I = 5X500X10-3 = 2.5W
(ii) Resistance = V/I = 5/500 X 10 -5 = 10 Ω
(iii) Energy consumed in four hrs.
= 2.5X4X60X60
= 10X3600 W.s.
= 3.6X104 J
Question. If a 12 V battery is connected to the arrangement of resistances given below, calculate:
a. the total effective resistance of the arrangement and
b. the total current flowing in the circuit.
Answer :
a. 10 W and 20 W are in series
Rs1 = 10 + 20 = 30 W
5 Ω and 25 Ω are in series
Rs2 = 5 + 25 = 30 W
30 W and 30 W are in parallel.
1/RP = 1/30 + 1/30 = 2/30
Rp = 15 Ω
(ii) I = V/RP = 12/15 = 0.8 A
Question. An electric bulb is rated at 200 V–100 W. What is its resistance? Five such bulbs bum for 4 hours. What is the electrical energy consumed? Calculate the cost, if the rate is 50 paise unit.
Answer : V = 200 V
P = 100W
(a) P = V2/R
R = (200)2/100 Ω = 40000/100 = 400 Ω
(b) Energy consumed in 4 hrs. by one bulb.
Question. Three 2 Ω resistors, A, B and C, are connected as shown in figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.
Answer :
Given maximum power of the network
P = 18W
Maximum current in 2 W resistor.
Resistor 2 W, 2 W are equal and in parallel combination so current of 3 A will be distributed equally.
I1 = I2 = 1.5 A
