Please see Chapter 12 Electricity Exam Questions Class 10 Science below. These important questions with solutions have been prepared based on the latest examination guidelines and syllabus issued by CBSE, NCERT, and KVS. We have provided Class 10 Science Questions and answers for all chapters in your NCERT Book for Class 10 Science. These solved problems for Electricity in Class 10 Science will help you to score more marks in upcoming examinations.
Exam Questions Chapter 12 Electricity Class 10 Science
ONE MARK QUESTIONS
Question: If the charge on an electron be 1.6 x10-19C, find the approximate number of electrons in 1 C.
Answer: 1.6×10-19C charge is of = 1 electron and
1 C charge is of =1/1.6×10-19 electron
No. of electrons = 6.25×1018
Question: What is the SI unit of electric potential?
Answer: Volt is the SI unit of electric potential.
Question: List any two factors on which resistance of a conductor depends.
Answer: Resistance of a conductor:
a. is directly proportional to its length
R ∝ p
b. is inversely proportional to its area of cross section.
R∝ 1/A
Combining (1) and (2), we get
R∝p/A
Question: Mention one reason why tungsten is used for making filament of electric lamp.
Answer: Tungsten is used for making filament because of its high melting point and low resistivity.
Question: In an electric circuit, state the relationship between the direction of conventional current and the direction of flow of electrons.
Answer: Electrons flows from negative terminal to positive terminal where as current flows from +ve terminal to -ve terminal in external circuit i.e. Conventional current and electrons flow are opposite to each other.
Question: Write SI unit of resistivity.
Answer: Ohm-m
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)t
E = Pxt = 60×1 J
E = 60 J
Question: Draw a schematic diagram of a circuit consisting of a cell of 1.5 V, 10 ohm resistor and 15 ohm resistor and a plug key all connected in series.
Answer: Schematic diagram is shown below.
Question: Why do we use copper and aluminium wire for transmission of electric current?
Answer: Copper and aluminium wires are used for electric transmission due to their low resistivity.
TWO MARKS QUESTIONS
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 t
Q = 4 A
= 12 = 3 A
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: 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: 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 Ω
V = 220 volt
V = IR or I= V/R
I=220\ 1100=1/5A
When heater is connected with the same source then
I=V/R=220/100=2.2A
Question: Calculate the number of electrons that would flow per second through the cross- section of a wire when 1 A current flows in it.
Answer:
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: List in a tabular form two differences between a voltmeter and an ammeter.
Answer:
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.
Question: (a) What material is used in making the filament of an electric bulb?
(b) Name the characteristics which make it suitable for this.
Answer: a. Tungsten is used in making filament.
b. Its high resistivity and high melting point.
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: 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 bulb is rated at 5.0 V, 100 mA. Calculate its (a) power and (b) resistance.
Answer:
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=0.2 A X 1 hr =0.2 X 60 X 60 A – s
=720 C
Question. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer.Q = 96000 C, t = 1 hr = 3600 s, V = 50 volt
H = V × I × t = V × Q = 50 × 96000 C =(50 × 96000)/1000 kJ = 4800 kJ
Question. An electric iron of resistance 20 W takes a current of 5 A. Calculate the heat developed in 30 s.
Answer.R = 20 Ω, I = 5 A, t = 30 s
H = I2 × R × t = 5 × 5 × 20 × 30 = 15000 J = 15 kJ
Question. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hours.
Answer.I = 5 A, V = 220 V, P = ?, E = ?, t = 2 hours
P = V × I = 220 V × 5 A = 1100 W
E = P × t = 1100 W × 2 h = 2200 W h = 2.2 kW h
Question. Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line, if the maximum allowable current is 5 A?
Answer.Total power = N × P where ‘N’ is the number of bulbs, ‘P’ power of each bulb.
Question. Find the resistance of bulb rated as 100 W at 250 V.
Answer.
Question. Name the commercial unit of energy. Convert it into Joules. What is other name of commercial unit of energy?
Answer.The commercial unit of electrical energy is kW h. (Kilowatt hour)
1 kW h = 1 kW × 1 hour = 1000 W × 3600 s = 3.6 × 106 J
Question. Find the minimum rating of fuse that can be safely used on a line on which two 1.1 kW rating electric geysers are to be run simultaneously. The supply voltage is 220 V.
Answer.
The rating of fuse that can be used is more than 5 A.
It is also called Board of Trade Unit (BOTU)
Question. An electric heater of resistance 8 W draws 15 A current from the supply mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Answer.
Question. A potential difference of 4 V derive a current of 3 A through a resistor. How much electrical energy will be converted into heat during 10 seconds?
Answer.
Question. An electric fan has a rating of 460 W on the 230 V mains line. What fuse should be fitted in the plug?
Answer.P = V × I ⇒ 460 = 230 × I ⇒ I = 2 Ampere
A 3A fuse should be fitted (It should be slightly more than 2A).
Question. What is the cost of running an AC with average power of 1000 W for 8 hours for 30 days. The cost of electric energy is `4.70 per kW h.
Answer.E = P × t = 1000 W × 8 × 30 = 240000 W h = 240 kW h
Cost of electric energy = 240 × 4.70 = rs 1128.
Question. Why do we get electric shock in damp conditions?
Answer.Water provides conducting path for a current to flow through the human body in damp conditions like bathroom. We get electric shock if we are bare footed. Wet body has low resistance, high current can easily pass through, leading to electric shock.
Question. Compare the power used in the 2 W resistor in each of the following circuits: (i) a 6 V battery in series with the 1 W and 2 W resistors, and (ii) a 4 V battery in parallel with 12 W and 2 W resistors.
Answer.
Question. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electrical mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer.
Question. (i) Explain what is the difference between direct current and alternating current. Write one important advantage of using alternating current.
(ii) An air conditioner of 2 kW is used in an electric circuit having a fuse rating of 10 A. If the potential difference of the supply is 220 V, will the fuse be able to withstand when the air conditioner is switched on? Justify your answer.
Answer.(i) The current whose direction gets reversed after every half cycle is called alternating current or A.C.
There is no change in the direction of D.C. D.C. is a uni directional current.
The most important advantage of using A.C. over D.C. is that in the A.C. mode electric power can be transmitted over long distances at a high voltage and low current with very little loss of power.
(ii) Here P = 2 kW = 2000 W, V = 220 Volt
P = VI, so the current, I = P/V =2000/220 = 9.09 A
As the current is 9.09 A below the rating of fuse, therefore the fuse will withstand, i.e. it will not blow off when A.C. is on.
Question. What is meant by ‘electrical resistance’ of a conductor? State how resistance of a conductor is effected when
(i) Low current passes through it for a short duration.
(ii) A heavy current passes through it for 30 seconds.
Answer.It is in opposite direction offered to the flow of electric current.
(i) No effect on resistance, low current, hence no appreciable rise in temperature so there no change in resistance.
(ii) Heavy current for 30 seconds may increase the temperature so resistance will increase.
Question. An electric kettle of 2 kW works for 2 hours daily. Calculate the
(a) Energy consumed in SI unit and commercial unit.
(b) Cost of running it in the month of June at a rate of ` 3.00 per unit.
Answer.(a) P = 2 kW, t = 2 h
E = P × t = 2 × 2 h = 4 kW h
(b) Total energy consumed per month = 4 kW h × 30 = 120 kW h
Total cost = 120 × 3 = rs 360
Question. An electric bulb is rated 220 V and 100 W, when it is operated at 110 V. What will be the power consumed?
Answer.
Question. An electric iron has a rating of 750 W, 220 V. Calculate the
(i) Current flowing through it and (ii) Its resistance when it is in use.
Answer.(i) P = 750 W, V = 220 V
P = V × I ⇒ 750 = 220 × I ⇒ I = 750/220
= 3.409 A
Question. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 4 W in series with a combination of two resistors (8 W each) in parallel and a voltmeter across parallel combination. Each of them dissipate maximum energy and can withstand a maximum power of 16 W without melting. Find the maximum current that can flow through the three resistors.
Answer.
Question. Two lamps, one rated 100 W at 220 V and the other 200 W at 220V are connected (i) in series and
(ii) in parallel to electric main supply of 220V. Find the current drawn in each case.
Answer.
Question. How many 40 W; 220 V rating lamps can be safely connected to a 220 V, 5 A line? Justify your answer.
Answer.
THREE MARKS QUESTIONS
Question: What is meant by “electrical resistance” of a conductor? State how resistance of a conductor is affected when
a. a low current passes through it for a short duration;
b. a heavy current passes through it for about 30 seconds.
Answer: Electrical resistance is the property of a conductor by virtue of which it opposes the flow of current through it. It is equal to the ratio of the potential difference applied across its ends to the current flowing through it.
R= V/I
a. When a low current is passed for a short duration, through a conductor, heat produced is almost negligible and hence no appreciable change in its resistance.
b. When heavy current is passed through the conductor for 30 s. Conductor may be get heated and its resistance and resistivity change.
Question: Name and define the SI unit of current. Calculate the number of electrons that flow through a conductor in 1 second to constitute a current of 1 ampere. (Charge on an electron = 1.6 X 10-19 coulomb)
Answer: SI unit of current is Ampere (A)
I=q/t
If q = 1 C, t = 1 s
then I = 1 A
If 1 C charge flows in 1 s in a conductor then magnitude of current is said to be 1 A.
q = n e
n= q/e = I X t/e
=1 A X 1 s / 1.6 X 10-19=100/16 X 1018
=6.25 X 1018
Question: Electric current flows through three lamps when arranged in (a) a series (b) a parallel. If the filament of one lamp breaks. Explain what happens to the other two lamps in both the cases.
Answer: a. In series combination if the filament of one lamp breaks then the circuit will be broken and hence other lamps stops glowing.
b. In parallel combination of lamps if the element of one lamp breaks then other two will continue to glow.
Question: Show how would you join three resistors, each of resistance 9 so that the equivalent resistance of the
combination is (i) 13.5 Ω (ii) 6 Ω ?
or
(a) Write Joule’s law of heating.
(b) Two bulbs, one rated 100 W; 220 V, and the other 60 W; 220 V are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line if the supply voltage is 220 V.
Answer:
Question: Study the V-I graph for a resistor as shown in the figure and prepare a table showing the values of I (in amperes) corresponding to four different values of V (in volts). Find the value of current for V = 10 volts. How can we determine the resistance of the resistor from this graph?
Answer: When V = 10 volt from the graph R =Δ V/Δ I=8-6/4-3 Ω =2 Ω
when V = 10 volt
then V = IR
or I = V/R=10/2 A= 5 A
Question: V-I graphs for two wires A and B are shown in the figure. If both the wires are made of the same material and are of equal thickness, which of the two is of more length? Give justification for your answer.
Answer: We know for identical wire more length more resistance and vice versa slope of wire A is more than B. Hence resistance of A is more and its length also.
Question: Find the number of electrons transferred between two points kept at a potential difference of 20 V if 40 J of work is done.
Answer:
Given: V = 20 Volt
W = 40 J
W = P X t
= V X I X t
=V X Q/t X t
or W = V X Q
40 = 20 X Q
or Q =2 C
n e = Q
n =Q/e=2/1.6 X 1019
1.25 X 1019
Question: (a) Why are copper or aluminium wires generally used for electrical transmission and distribution purposes?
(b) Two wires, one of copper and other of manganin, have equal lengths and equal resistances. Which wire is thicker? Given that resistivity of copper is lower than that of manganin.
Answer: a. Copper or aluminium wires are used for transmission and distribution of electricity due to their low resistivity and high conductivity.
b. We know that R=p l/A
p∝A
Thicker the wire, more the resistivity. The resistivity of manganin is more than copper. So manganin wire is thicker than copper.
Question: (a) State Ohm’s law. Express it mathematically.
(b) Write symbols used in electric circuits to represent:
(i) variable resistance.
(ii) voltmeter.
(c) An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, what will be the power consumed?
Answer: a. Ohm’s Law states that if the physical conditions of a conductor are kept constant then current passing through a conductor is directly proportional to the potential difference across its ends.
Question: (a) Define the term ‘volt’.
(b) State the relation between work, charge and potential difference for an electric circuit.
Calculate the potential difference between the two terminals of a battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to the other.
Answer: a. Potential difference b/w two points in an electric field is said to be 1 volt if the amount of work done in bringing a unit positive charge from one point to another point is 1 J.
b. Given: W = 100 J, Q = 20 C, V = ?
As V=W/ Q ⇒ V=100/20 J C-1
V = 5 J C-1
V = 5 Volt.
FIVE MARKS QUESTIONS
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 symbols. 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: By changing the position of ammeter, measure the current in it. Every time the magnitude of current is found same i.e., in series combination the current in every part of the circuit, remains same.
Precaution:
a. Connect the ammeter in series with battery.
b. All connection must be tight.
c. Current must not be greater than the range of ammeter.
Question: a. What is meant by saying that the potential difference between two points is 1 volt?
b. Why does the connecting cord of an electrical heater not glow while the heating element does?
c. Electrical resistivity of some substance at 20°C are given below:
Answer the following questions in relation to them:
i. Among silver and copper which one is better conductor? Why?
ii. Which material would you advise to use in electrical heating device? Why?
Answer: a. If the amount of work done in bringing a unit positive charge from one point to another is 1 J in an electric field then potential difference b/w the two points is said to be 1 volt.
1V=1J/1C
b. Connected cord of an electrical heater does not glow because this is of copper metal, which has low resistivity and good conductivity. Heating element glows due to its high resistivity or poor conductivity.
c. (i) The resistivity of silver is lesser than copper so it is a good conductor.
(ii) The resistivity of nichrome is maximum in the given table so nichrome will be advised to use in electrical heating device.
Question: (a) Name an instrument that measures electric current in a circuit. Define unit of electric current.
(b) What are the following symbols mean in an electric circuit?
(c) Draw a closed circuit diagram consisting of 0.5 m long nichrome wire XY, an , ammeter, a voltmeter, four cells of 1.5 V and a plug key.
Answer: a. Ammeter.
If IC charge flows in an electric circuit is 1 s then the current is said to be 1 A.
b. (i) Rheostat (ii) Closed key
Question: (a) Derive the formula for the calculation of work done when current flows through a resistor.
(b) One electric bulb is rated 40 W and 240 V and other 25 W and 240 V. Which bulb has higher resistance and how many times?
Answer: a. Let R is the resistance of the resistor in which I
current is passed for a time t by applying a p.d. ( V ) across the resistor.
The work done in bringing a charge Q from one end to another of the conductor
Question: The value of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below :
Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor.
Answer:
Question: (a) Name an instrument that measures potential difference between two points in a circuit. Define the unit of potential difference in terms of SI unit of charge and work. Draw the circuit symbols for (i) variable resistor, (ii) a plug key which is closed one.
(b) Two electric circuits I and II are shown below “
(i) Which of the two circuits has more resistance?
(ii) Through which circuit more current passes?
(iii) In which circuit, the potential difference across each resistor is equal?
(iv) If R1 > R2 > R3 in which circuit more heat will be produced in R1 as compared to other two resistors?
Answer: a. Voltmeter The amount of work done in bringing a unit positive charge from one point to another in an electric field is said to be potential difference
V=W/Q
b. (i) In series combination resistance is more than parallel combination.
(ii) Lesser the resistance more the current in circuit i.e., in parallel (II) current is max.
(iii) In parallel combination (II)
(iv) More heat in (I) across R1.
Question: When a high resistance voltmeter is connected directly across an electric bulb, its reading is 2 V. An electric cell is sending the current of 0.4 ampere (measured by an ammeter) in the electric circuit.
a. Draw the circuit.
b. Find the resistance of the electric bulb,
c. State the law that is applied for making these calculation. If a graph is plotted between V and I, show the nature of the graph obtained.
Answer: a.
b. V = 2 Volt, I = 0.4 A
R =V/I =2/0.4
ohm or R = 5 ohm
c. Ohm’s law: If the physical conditions of a conductor is kept constant then current in the circuit is directly proportional to the potential difference applied across the ends of the conductor.
Since V ∝ I
Hence graph between V and I is a straight line for a conductor. Which passes through the origin O of the graph.
Question: (a) State the commercial unit of electric energy and find its relation with its SI unit.
(b) The current through a resistor is made three times its initial value. Calculate how it will affect the heat produced in the resistor.
(c) Find the increase in the amount of heat generated in a conductor if another conductor of double resistance is connected in the circuit keeping all other factors unchanged.
Answer: a. Commercial unit of electric energy = kWh
1 kWh = 3. 6 X 1 06 J
b. Initial heat generated in the
resistor = I 2 Rt
H 1 = I 2 Rt
when current is made three times i.e. 3 I now heat generated
H 2 = (3 I) 2 Rt
H 2 = 9 I 2 Rt
In later case, heat generated is 9 times the initial heat generated.
c. If another conductor of 2 R is connected in series
then total resistance = R + 2 R = 3 R.
Now heat generated H = I 2 (3 R)t
H = 3I 2Rt
In this case, heat generated is three times.
Question: Draw a schematic diagram of an electric circuit (in the “on” position) consisting of a battery of five cells of 2 V each, a 5 Ω resistor, a 8 Ω resistor, a 12 Ω resistor and a plug key, all connected in series. An ammeter is put in the circuit to measure the electric current through the resistors and a voltmeter is connected so as to measure the potential difference across the 12 Ω W resistor.
Calculate the reading shown by the: (a) ammeter (b) voltmeter in the below electric circuit.
Answer: Resistors of 5 Ω, 8 Ω , 12 Ω all the connected in series. Hence,
RS = R1+ R2+ R3
= (5 + 8 + 12) Ω
RS = 25 Ω
V=IR
I =V/R=10/25 A
I=0.4 A
a. Ammeter reading is 0.4 A.
b. V = IR
Potential drop across 12 W resistance V = 0.4 X 12 = 4.8 Volt.
Question: (a) Explain how does a cell maintain current in a circuit.
(b) In the circuit given below the resistance of the path xTy = 2 W and that of xZy = 6 Ω
(i) Find the equivalent resistance between x and y.
(ii) Find the current in the main circuit.
(iii) Calculate the current that flows through the path xTy and xZy.
Answer: a. Potential difference b/w the two terminals of a cell is produced due to chemical reaction in the cell. This potential difference maintains a current in the circuit.
b. (i) The resistance b/w XZY and XTY are in parallel combination. Their equivalent resistance.
Question: (a) Derive an expression to find the equivalent resistance of three resistors connected in series.
Also, draw the schematic diagram of the circuit.
(b) Find the equivalent resistance of the following circuit:
Answer: a.
If the equivalent resistance is RS
then, V = IRS
But, V = V1+ V2+ V3
RS = R1+ R2+ R3
b.
1/RP=1/R1+1/R2+1/R3
=1/6+1/10+1/15=5+3+2/30
=10/30
R P=3 Ω
Question: Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit.
Calculate the resistance of 50 m length of wire of cross-sectional area 0.01 square mm and of resistivity 5X10-8 Ωm .
Answer: Resistance is the opposition offered in the path of flow of current by the atoms or molecules of a conductor. Factors affecting resistances:
a. length R∝ l
b. area of cross-section R∝1/A
c. nature of material.
Rheostat is used to change the current in the circuit without changing the voltage source.
Given: l = 50 m
A = 0.01 mm2 = 0.01X10-6 m2
P=5 X108 ΩM
As R=P l/A=5X108X 50/0.01X10-6
R=250Ω
Question: (a) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected to battery, ammeter, voltmeter and key. Draw suitable circuit diagram. Obtain an expression for the effective resistance of the combination of resistors in parallel.
(b) Why are electric bulbs filled with chemically inactive nitrogen or argon?
(c) What is meant by the statement that the rating of a fuse in a circuit is 5 A?
Answer:
a. Let equivalent (effective) resistance is R then I =V/R
Similarly, I1 V /R1
I2 =V/R2and I3=V/R3
But I = I1+ I2+ I3
or V/R=V/R1+V/R2+V/R3
1/R=1/R1+1/R2+1/R3
b. To prevent oxidising the filament due to high temperature.
c. The current in the fuse wire must not exceed 5 A otherwise it will melt.
Question:(a) Calculate the resistance of the wire using graph.
(b) How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?
(c) Define electric power, Derive relation between power, potential difference and resistance.
Answer:
Question: (a) Name and state the law that gives relationship between the current through a conductor and the potential difference across its two terminals. Also, express this law mathematically.
(b) Draw the V-I graph for this law. Justify your answer.
(c) Write the name and use of the circuit components
whose symbols are given below.
Answer: a. The law is Ohm’s law.
If the physical conditions of a conductor is kept constant then current through it is directly proportional to the potential difference applied across it.
V ∝ I or V = RI
b. Since V∝ I so a graph b/w V and I is a straight line.
c. (i) Symbol is of variable resistor and it is used to regulate the current.
(ii) Plug key is closed. When plug key is closed current flows through the circuit.
Question: Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 Ω resistor, a 10 W resistor and a 15 Ω resistor, an ammeter and a plug key; all connected in series. Also, connect a voltmeter to record the potential difference across the 15 Ω resistor and calculate:
a. the electric current passing through the above circuit and
b. potential difference across 5 W resistor when the key is closed.
Answer:
a. Net effective resistance of the circuit.
R (5 + 10 + 15)Ω = 30 Ω
Current in the circuit
I=ne/R=5X2/30A=1/3 A
I = 0.33 A
b. P.D. across 5 W resistor V = IR
V = 0.33X5 = 1.65 Volt.
V = 1.65 Volt.
Question: (a) In the circuit shown 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 cross-section 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 1V”. Explain the statement.
Answer: a. (i) The resistance of two times long wire also becomes two times so current decreases in the circuit.
(ii) If area of the nichrome wire is doubled then its resistance decreases and hence current increases in the circuit.
b. Potential difference b/w A and B is 1 volt means that 1 J of work is to be done in moving a unit positive charge (+ 1C) from point A to B.
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
= 200X5 = 1000Wh
Energy consumed by electric press per day
= 500WX4 hr
= 2000Wh
Total energy consumed per day
=(1000 + 2000)Wh
= 3000Wh = 3 kWh
Total energy consumed in 30 days
= 3X30 kWh
= 90 kWh
Question:(a) Define potential difference between two points in a conductor.
(b) Name the instrument used to measure the potential difference in a circuit. How is it connected?
(c) A current of 2 A passes through a circuit for 1 minute. If potential difference between the terminals of the circuit is 3 V, what is the work done in transferring the charges?
Answer: a. Electric potential is the amount of work done in bringing a unit positive charge from one point to another.
b. Voltmeter. It is connected in parallel in the circuit.
c. I = 2A, t = 1 min. = 60 s, V = 3 V
W = VQ = V(It)
W = 3X2X60 J
W = 360 J
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
I Volt/1 Ampere =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) What is an electric circuit?
(b) Calculate the number of electron that flow per second to constitute a current of one ampere.
Charge on an electron is 1.6X10-19 C.
(c) Draw an electric circuit for studying Ohm’s law. Label the circuit component used to measure electric current and potential difference.
Answer: a. A continuous path in which current can flow when switch i6 plugged in.
b. Q = It or ne = It ⇒n =I.t/e
Given: I = 1A, t =1s
n=1X1/1.6X10-19
n = 6.25X1018 electrons
c. K–Plug key, (A)–Ammeter, (V)–Voltmeter, Rh– Rheostat
Question: (a) Name an instrument that measures electric current in a circuit. Define the unit of electric current.
(b) What do the following symbols mean in circuit diagrams?
(c) An electric circuit consisting of a 0.5 m long nichrome wire XY, an ammeter, a voltmeter, four cells of 1.5 V each and a plug key was set up.
(i) Draw a diagram of this electric circuit to study the relation between the potential difference maintained between the points X and Y and the electric current flowing through XY.
(ii) Following graph was plotted between V and I values:
What would be the values of V/I ratios when the potential difference is 0.8 V, 1.2 V and 1.6 V respectively? What conclusion do you draw from these values?
Answer: a. Ammeter: Ampere is the unit of current. If one coulomb charge flows in a circuit in 1 s then the current in the circuit will be 1 Ampere. (A)
b. (i) Variable resistor (ii) closed key
The graph b/w V and I is a straight line.
R = V/I = 1.6-0/0.6-0 =1.6/0.6 =2.67 Ω
Question. An electric circuit two resistors of 20 W and a conductor of resistance 4W are connected to a 6 V vattery as shown in the figure. Calculate:
(a) The total resistance of the circuit,
(b) the current through the circuit,
(c) the potential difference across the (i) electric lamp and (ii) conductor, and
(d) power of the lamp.
Answer.
Question. Explain the following:
(i) Why is tungsten used almost exclusively for filament of electric lamps?
(ii) Why are the conductors of electric heating devices, such as bread-toaster and electric iron, made of an alloy rather than a pure metal?
(iii) Why is series arrangement not used for domestic circuits?
(iv) How does the resistance of a wire vary with its area of cross-section?
(v) Why are copper and aluminium wires usually used for the transmission of electric current?
Answer.(i) It has high resistance and high melting point. So it does not melt when current is passed through it.
(ii) Alloys have more resistivity and higher melting point.
(iii) It increases the resistance and current decreases. Also if one component fails to work, others will also not work
(iv) Resistance is inversely proportional to the area of cross section.
(v) It is because Cu and Al have low resistivity and it allows current to flow.
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 fuse that are respectively rated at 1 A, 3 A and 5 A can be used for it.
Answer.(a) Electric power is the rate of doing work by an energy source or the rate at which the electrical energy dissipated or consumed per unit time in the electric circuit.
(b) It means the maximum current will flow through it is only 2 A. Fuse wire will melt if the current
flowing through it exceeds 2 A value.
(c) Given: P = 1 kW = 1000 W, V = 220 V
Current drawn, I = P/V
1000/220=50/11= 4.54 A
To run electric iron of 1 kW efficiently, rated fuse of 5 A should be used.
Question. (a) Write two points of differences between electric energy and electric power.
(b) 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 the units of Joule.
Answer.(a) Difference between electric energy and electric power:
(b) For the same applied voltage, P ∝ 1/R
i.e. smaller the power of electrical device, higher is its electrical resistance.
So a 40 W lamp has higher electrical resistance than a 60 W lamp.
(c) Kilowatt hour – Commercial unit of electrical energy.
1 kW h = 1000 W h = 1000 J/s × 3600 s = 3600000 J = 3.6 × 106 J
Question. Two identical wires, one of nichrome and other of copper are connected in series and a current (i) is passed through them. State the change observed in the temperature of the two wires. Justify your answer. State the law which explains the above observations.
Answer.The resistivity of nichrome is more than that of copper, so its resistance is also high. Therefore large amount of heat is produced in the nichrome wire for the same current passed as compared to that of the copper wire. Accordingly greater change in temperature is observed in case of nichrome wire. This can be explained by Joule’s law of heating.
Joule’s law of heating states that the amount of heat produced in a conductor is:
(i) Directly proportional to the square of current flowing through it, i.e.
H ∝ I2
(ii) Directly proportional to the resistance offered by the conductor to the flow current, i.e.
H ∝ R
(iii) Directly proportional to the time for which current is flowing through the conductor, i.e.
H ∝ t
Combining these, we get, H ∝ I2Rt, or H = KI2Rt
where K is a proportionality constant and in SI units, it is equal to unity.
Question. (a) State Ohm’s law. Derive the relation and give graphical representation for it.
(b) An electric oven rated at 500 W is connected to a 220 V line and used for 2 hours daily. Calculate the cost of electric energy per month at the rate of ` 5 per kW h.
Answer.(a) Linear equation, hence a straight line passing through the origin is obtained as follows:
Ohm’s law states that potential difference across the given metallic wire in an electric circuit is directly proportional to the electric current flowing through it, if the temperature remains constant.
Mathematically, V ∝ I.
or V/I = Constant
or V = IR (where R is a constant known as the constant of proportionality).
(b) Energy consumed per day = 1 kW h.
P = V × I, E = P × t
E = V × I × t = 220 × 2.27 × 2 = 1 kWh
So, I = P/V = 500/220
= 2.27 A (P = V × I,I=500/220)
Cost of electric energy for 30 days = 1 × 5 × 30 = rs 150.00
Question. In the given circuit, A, B, C and D are four lamps connected with a battery of 60V.Analyse the circuit to answer the following questions.
(i) What kind of combination are the lamps arranged in (series or parallel)?
(ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?
(iii) Explain with proper calculations which lamp glows the brightest?
(iv) Find out the total resistance of the circuit.
Answer.(i) The lamps are in parallel.
(ii) Advantages:
If one lamp is faulty, it will not affect the working of the other lamps.
They will also be using the full potential of the battery as they are connected in parallel.
(iii) The lamp with the highest power will glow the brightest.
P = VI
In this case, all the bulbs have the same voltage. But lamp C has the highest current.
Hence, for lamp C, P = 5 × 60 Watt = 300 W. (the maximum).
(iv) The total current in the circuit = (3 + 4 + 5 + 3) A = 15A
The Voltage = 60V
V = IR and hence R = V/I
R = 60/15 W = 4 W
Question. (a) What is the function of fuse wire in an electric circuit?
(b) What would be the rating of the fuse for an electric kettle which is operated at 220 V and consumes 500 W power?
(c) How is the SI unit of electric energy related to its commercial unit?
Answer.(a) Fuse wire protects the electrical circuits from over voltage and high current.
(b) 2.2 A flows through the circuit, fuse should be rated 3 A.
(c) 1 kW h = 3.6 × 106 J
Question. (i) Consider a conductor of resistance ‘R’, length ‘L’, thickness ‘d’ and resistivity ‘ρ’.
Now this conductor is cut into four equal parts. What will be the new resistivity of each of these parts? Why?
(ii) Find the resistance if all of these parts are connected in: (a) Parallel (b) Series
(iii) Out of the combinations of resistors mentioned above in the previous part, for a given voltage which combination will consume more power and why?
Answer.
(i) Resistivity will not change as it depends on the nature of the material of the conductor.
(ii) The length of each part becomes L/4. While ρ, A remain constant. R = ρL/A.
Resistance of each part = Rpart = (ρL/4)/A = R/4.
(iii) P = V2/R.
If Reqv is less, power consumed will be more.
In the given case, Reqv is lesser in the parallel connection and the power consumed will be more.
