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# Electricity

### NCERT Questions

##### Question1. What does an electric circuit mean?

Answer:A continuous and closed path of an electric current is called an electric circuit

##### Question2. Define the unit of current?

Answer:One ampere is the current flowing through a circuit, when a charge of one coulomb flows through it in one second

##### Question3. Calculate the number of electrons constituting one coulomb of charge?

Answer:Total charge, Q = n x e

##### Question4. Name a device that helps to maintain a potential difference across a conductor?

Answer:With the help of a cell or a battery

##### Question5. What is meant by saying that potential difference between two points is 1 volt?

Answer:It means that 1 joule work is done in moving 1 coulomb of positive charge from one point to the other in an elecfric field

##### Question7. On what factors does the resistance of a conductor depend?

Answer:Resistance of a conductor depends on its material, length and area of cross-section

##### Question8. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer:According to Ohm law, we have: I ? V or V ? I Since the potential difference (V) decreases to half, i.e. ? V2?Ie Hence, the current flowing through the component will become half

##### Question9. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer:This is because of the high resistivity and high melting point of the alloy. Also, they do not oxidise when red hot

##### Question11. Will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?

Answer:The current will flow more easily through a wire which have comparatively less resistance. According to the relation R?1A, the thin wire has less resistance. So, the current will flow more easily through thin wire

##### Question12. Judge the equivalent resistance when the following are connected in parallel?
• 1 Ω and 106 Ω
• 1 Ω and 103 Ω, and 106 Ω

• Approximately but less than 1 Ω, because in a parallel combination of resistors, the equivalent resistance is less than the least resistance
• Same as in previous part of the question
##### Question13. Redraw the circuit of previous question putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer:Total resistance, Rs = 5 + 8 + 12 = 25 Ω

##### Question14. An electric lamp of resistance 100 12, a toaster of resistance 50 11 and a water filter of resistance 500 12 are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer:Given: Resistance of electric lamp, R1 = 100Ω Resistance of toaster, R2 = 50Ω Resistance of water filter, R3 = 500Ω Voltage, V = 220 V Let Rp be the resistance of electric iron. Then

##### Question15. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?
ans:
• The highest resistance can be secured by connecting all the four coils in series. ? Equivalent resistance of series combination, Rs= 4 + 8 + 12+ 24 = 48 Ω Hence, the highest resistance that can be secured by given coils is 48 Ω
• The lowest resistance can be secured by connecting all the four coils in parallel. ? Equivalent resistance of parallel combination, 1Rp=14+18+112+124=12 or Rp = 2 Ω Hence, the lowest resistance that can be secured by given coils is 2 Ω
##### Question16. Why does the cord of an electric heater not glow while the heating element does?

Answer:The resistance of metal used in cord has much lower than that its element. So the heat produced by the element is much more than that produced by the cord while passing the current

##### Question17. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V?

Answer:Given, Q = 96000 C, V= 50 volt,
t = 1 hour = 60 x 60 = 3600 s
? Heat generated, H = VIt
= VQ = 50 x 96000
= 4800000 = 4.8 x 106 J

##### Question18. An electric iron of resistance 20 12 takes a current of 5 A. Calculate the heat developed in 30 s?

Answer:Heat developed is equal to the work done (W).
Given: R = 20 Ω, I = 5 A and t = 30 s
? W = I2Rt = (5)2 x 20 x 30
= 25 x 20 x 30 = 15,000 J
Hence, the heat developed by electric iron is 15,000 J

##### Question19. What determines the rate at which energy is delivered by a current?

Answer:Electric power determines the rate at which energy is delivered by a current

##### Question20. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hr?

Answer:Given: Current, I = 5 A, Voltage, V = 220 V and t= 2 h
Electric power, P=VI = 220 x 5 = 1100 W
? Energy consumed = Pt = 1100 x 7200
= 7920000 = 792 x 104 J

##### Question21. What are the advantages of connecting devices in parallel with the battery instead of connecting them in series?

• When electrical devices are connected in parallel, each is operated at the same voltage and if one of them goes off, the others are not affected. But when electrical devices are connected in series, this does not happen because the circuit is broken when any one of the devices fails
• In parallel connection, the current divides itself through different devices. This is helpful when each device has different resistance and requires different current to operate. But in series connection this does not happen because the same current flows through all the devices
##### Question22. How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of?
• 4 Ω and
• 1 Ω

• In order to get a resistance of 4 Ω, we can connect resistors of 3 Ω and 6 Ω in parallel and this parallel combination in series with 2 Ω resistor
• In order to get 1 Ω resistance, we can connect all the three resistors in parallel
• 1:2
• 2:1
• 1:4
• 4:1
ans:
• 1:4
• 100w
• 75w
• 50w
• 25w

• 25w

##### Question7. thevaluedofcurrentiflowinginagivenresistorforthecorrespondingvaluesofpotentialdifferencevacrosstheresistoraregivenbelow?
1.6
i(amperes) v(volts)
0.5 1.0 2.0 3.0 4.0
3.4 13.2

plotagraphbetweenvandiandcalculatetheresistanceoftheresistor.

##### Question9. abatteryof9visconnectedinserieswithresistorsof0.2Ω,0.3Ω,0.4Ω,0.5Ωand12Ω,respectively.howmuchcurrentwouldflowthroughthe12Ωresistor?

applyingtheformular=r1+r2+r3+r4+rs,weget
r=0.2+0.3+0.4+0.5+12=13.4Ω
byohmlaw,wehave
v=rior
or9=13.4xi
ori=913.4=0.67a
hence,thecurrentflowsthroughthe12Ωresistoris0.67a

##### Question10. howmany176Ωresistors(inparallel)arerequiredtocarry5aona220vline?

resistanceofeachresistor,r=176Ω
numberofresistors=n,weknowthatv=rpi

##### Question11. showhowyouwouldconnectthreeresistorseachofresistance6ft,sothatthecombinationhasaresistanceof?
• 9Ω,and
• 4Ω

• inordertogetaresistanceof9ftfromthegiventhreeresistors,tworesistorscanbeconnectedinparallelandthisparallelcombinationinserieswiththethirdone
• inordertogetaresistanceof4ftfromthegiventhreeresistors,tworesistorscanbeconnectedinseriesandthisseriesinparallelwiththethirdone
##### Question12. severalelectricbulbsdesignedtobeusedona220velectricsupplyline,arerated10w.howmanylampscanbeconnectedinparallelwitheachotheracrossthetwowiresof220vline,ifthemaximumallowablecurrentis5a?

hence,110bulbscanbeconnectedisparallel.

##### Question14. comparethepowerusedinthe2Ωresistorineachofthefollowingcircuits?
• a6vbatteryinserieswith1Ωand2Ωresistors,and
• a4vbatteryinparallelwith12Ωand2Ωresistors

• given:voltage,v=6vequivalentresistance,r=1+2=3Ω
• sincethebatteryisinparallelwith12Ωand2Ωresistors ?potentialdifferenceacross2Ωresistor

##### Question18. explainthefollowing?
• whyisthetungstenusedalmostexclusivelyforfilamentofelectriclamps
• whyistheseriesarrangementnotusedfordomesticcircuits
• howdoestheresistanceofawirevarywithitsareaofcross-section
• whyarecopperandaluminiumwiresusuallyemployedforelectricitytransmission

• meltingpointoftungstenisveryhigh.soitdoesnotoxidise(burn)athightemperatures.secondly,theresistivityoftungstenishigh.sothinwireoftungstenwhichhashighresistancewhichproducesmoreheatenoughtoemitlight
• thisisbecauseofthehighresistivityandhighmeltingpointofthealloy.also,theydonotoxidisewhenredhot
• domesticcircuitsarenotconnectedinseriesbecausetheydonotworkatthesamevoltageandalsoifthereisashort-circuitinginonedistributioncircuit,itsfusewillnotblowandothercircuitswillalsobeaffected
• largertheareaofcross-sectionmeansalargenumberoffreeelectronscanmoveintheconductor,therebydecreasingtheresistance
• copperandaluminiumwiresaregoodconductorsofelectricity.sotheyhavelowresistivity.theyofferlowoppositiontotheflowofcurrentfromonepointtotheother,henceemployedforelectricity

Important Question

Important Questions of Electricity Class 10 Science Chapter 12
##### Question 1. A current of 10 A flows through a conductor for two minutes?Year of Question :(2012)
• (i) Calculate the amount of charge passed through any area of cross section of the conductor
• (ii) If the charge of an electron is 1.6 × 10-19 C, then calculate the total number of electrons flowing. (Board Term I, 2013)

Answer: Given that: I = 10 A, t = 2 min = 2 × 60 s = 120 s

• (i) Amount of charge Q passed through any area of cross-section is given by I = Qt or Q = I × t ? Q = (10 × 120) A s = 1200 C
• (ii) Since, Q = ne where n is the total number of electrons flowing and e is the charge on one electron ? 1200 = n × 1.6 × 10-19 or n = 12001.6×10?19 = 7.5 × 1021
##### Question 2. Define electric current?Year of Question :(2017)

Answer: Electric current is the amount of charge flowing through a particular area in unit time

##### Question 3. Define one ampere?Year of Question :(2015)

Answer: One ampere is constituted by the flow of one coulomb of charge per second. 1 A = 1 C s-1

##### Question 4. Name a device that you can use to maintain a potential difference between the ends of a conductor. Explain the process by which this device does so?Year of Question :(2015)

Answer: A cell or a battery can be used to maintain a potential difference between the ends of a conductor. The chemical reaction within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When it is connected to a conductor, it produces electric current and, maintain the potential difference across the ends of the conductor

##### Question 5. Draw the symbols of commonly used components in electric circuit diagrams for?Year of Question :(2012)
• (i) An electric cell
• (ii) Open plug key
• (iii) Wires crossing without connection
• (iv) Variable resistor
• (v) Battery
• (vi) Electric bulb
• (vii) Resistance (4/5, Board Term 1,2017)

##### Question 6. A student plots V-I graphs for three samples of nichrome wire with resistances R1, R2 and R3. Choose from the following the statements that holds true for this graph?Year of Question :(2014)
• (a) R1 = R2 = R3
• (b) R1 > R2 > R3
• (c) R3 > R2 > R1
• (d) R2 > R1 > R3

• (d) : The inverse of the slope of I-V graph gives the resistance of the material. Here the slope of -Rj is highest. Thus, R2 > R1 > R3
##### Question 7. State Ohms law?Year of Question :(2019)

Answer: It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically, V ? I V = RI where R is resistance of the conductor

##### Question 8. A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph?Year of Question :(2020)

Answer: As graph is a straight line, so it is clear from the graph that V ? I. The shape of the graph obtained by plotting potential difference applied across conductor against the current flowing v. llmuigh il will be a straight line. According to ohms law, V = IR or R = VI So, the slope of V-/ graph at any point represents the resistance of the given conductor

##### Question 9. 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 V (in volts). Find the value of current for V = 10 volts. How can we determine the resistance of the resistor from this graph?Year of Question :(2012)

Answer: Since, the graph is straight line so we can either extrapolate the data or simply mark the value from graph as shown in figure

Current, I(A) Voltage, V(V)
0 0
1 2
2 4
3 6
4 8
Hence, the value of current for V = 10 volts is 5 amperes (or 5 A). From Ohms law, V = IR, We can write, R = VI At any point on the graph, resistance is the ratio of values of V and I. Since, the given graph is straight line (ohmic conductor) so, the slope of graph will also give the resistance of the resistor R = 10V5A = 2? Alternately, R = (8?2)V(4?1)A = 6V3A = 2 ?

##### Question 10. V-I graph for a conductor is as shown in the figure ?Year of Question :(2012)
• (i) What do you infer from this graph
• (ii) State the law expressed here. (Board Term I, 2014)

• (i) As graph is a straight line, so it is clear from the graph that V ? I. The shape of the graph obtained by plotting potential difference applied across conductor against the current flowing v. llmuigh il will be a straight line. According to ohms law, V = IR or R = VI So, the slope of V-/ graph at any point represents the resistance of the given conductor
• (ii) It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically, V ? I V = RI where R is resistance of the conductor
##### Question 11. State Ohms law. Draw a labelled circuit diagram to verify this law in the laboratory. If you draw a graph between the potential difference and current flowing through a metallic conductor, what kind of curve will you get? Explain how would you use this graph to determine the resistance of the conductor?Year of Question :(2016)

Answer: As graph is a straight line, so it is clear from the graph that V ? I. The shape of the graph obtained by plotting potential difference applied across conductor against the current flowing v. llmuigh il will be a straight line. According to ohms law, V = IR or R = VI So, the slope of V-/ graph at any point represents the resistance of the given conductor

##### Question 12. State and explain Ohms law. Define resistance and give its SI unit. What is meant by 1 ohm resistance? Draw V-I graph for an ohmic conductor and list its two important features?Year of Question :(2012)

Answer: Resistance : It is ihe properly of a conductor lo resist the How of charges through it. Its SI unit is ohm (?). If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 ohm (1 ?). lvolt 1 ohm = 1volt1ampere V-I graph for an ohmic conductor can be drawn as given in figure. Important feature of V-I graph are: (i) It is a straight line passing through origin. (ii) Slope of V-I graph gives the value of resistance of conductor slope = R = VI

##### Question 13. Assertion (A) : The metals and alloys are good conductors of electricity. Reason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity?Year of Question :(2012)
• (a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A)
• (b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
• (c) (A) is true, but (R) is false.
• (d) (A) is false, but (R) is true. (2020)

• (c) : Metals and alloys are good conductors of electricity. Bronze is an alloy of copper and tin which are metals and thus is a good conductor of electricity
##### Question 14. A cylindrical conductor of length l and uniform area of cross section A has resistance R. The area of cross section of another conductor of same material and same resistance but of length 2l is ?Year of Question :(2020)
• (a) A2
• (b) 3A2
• (c) 2A
• (d) 3A

• (c) : The resistance of a conductor of length!, and area of cross section, A is R = ?lA where ? is the resistivity of the material. Now for the conductor of length 21, area of cross-section A and resistivity ?. R = ?lA = ?2lA But given, R = R ? ?lA = ?2lA or A = 2A
##### Question 15. Assertion (A) : Alloys are commonly used in electrical heating devices like electric iron and heater. Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points then their constituent metals?Year of Question :(2012)
• (a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A)
• (b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A)
• (c) (A) is true, but (R) is false.
• (d) (A) is false, but (R) is true. (2020)

• (a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A)
##### Question 16. How is the resistivity of alloys compared with those of pure metals from which they may have been formed?Year of Question :(2017)

Answer: The resistivity of an alloy is generally higher than that of its constituent metals

##### Question 17.
• (i) List three factors on which the resistance of a conductor depends
• (ii) Write the SI unit of resistivity. (Board Term 1, 2015)

• (i) Resistance of a conductor depends upon the following factors
• (1) Length of the conductor : (Treater the length (I) of the conductor more will be the resistance (R). R ? I
• (2) Area ol cross section of the conductor: (Ireater the cross-sectional area of the conductor, less will be the resistance. R ? 1A
• (3) Nature of conductor.
• (ii) SI unit of resistivity is ? m.
##### Question 18. Calculate the resistance of a metal wire of length 2m and area of cross section 1.55 × 106 m², if the resistivity of the metal be 2.8 × 10-8 ?m?Year of Question :(2013)

Answer: For the given metal wire, length, l = 2 m area of cross-section, A = 1.55 × 10-6 m² resistivity of the metal, p = 2.8 × 10-8 ? m Since, resistance, R = ?lA So R = (2.8×10?8×21.55×10?6)? = 5.61.55 × 10-2 ? = 3.6 × 10-2? or R = 0.036?

##### Question 19.
• (a) List the factors on which the resistance of a conductor in the shape of a wire depends.
• (b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason.
• (c) Why are alloys commonly used in electrical heating devices ? Give reason. (2018)

• (i) Resistance of a conductor depends upon the following factors
• (1) Length of the conductor : (Treater the length (I) of the conductor more will be the resistance (R). R ? I
• (2) Area ol cross section of the conductor: (Ireater the cross-sectional area of the conductor, less will be the resistance. R ? 1A
• (3) Nature of conductor.
• (ii) SI unit of resistivity is ? m.
• (b) Metal have very low resistivity and hence they are good conductors of electricity.
• Whereas glass has very high resistivity so glass is a bad conductor of electricity
• (c) Alloys are commonly used in electrical heating devices due to the following reasons
• (i) Alloys have higher resistivity than metals
• (ii) Alloys do not get oxidised or burn readily
• ##### Question 20. Calculate the resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance 20 ohms?Year of Question :(20127

Answer: We are given, the length of wire, l = 1 m, radius of wire, r = 0.01 cm = 1 × 10-4 m and resistance, R = 20? As we know, R = ?lA, where ? is resistivity of the material of the wire. ? 20?.= ?l?r2 = ?1 m3.14×(10?4)2 m2 ? ? = 6.28 × 10-7 ? m

##### Question 21. A copper wire has diameter 0.5 mm and resistivity 1.6 × 10-8 ? m. Calculate the length of this wire to make it resistance 100 ?. How much does the resistance change if the diameter is doubled without changing its length?Year of Question :(2015)

Answer: Given; resistivity of copper = 1.6 × 10-8 ? m, diameter of wire, d = 0.5 mm and resistance of wire, R = 100 ? Radius of wire, r = d2 = 0.52 mm = 0.25 mm = 2.5 × 10-4 m Area of cross-section of wire, A = nr² ? A = 3.14 × (2.5 × 10-4)² = 1.9625 × 10-7 m² = 1.9 × 10-7 m² As, R = ?lA ? 100 ? = 1.6×10?8?m×l1.9×10?7 m2 l = 1200 m If diameter is doubled (d = 2d), then the area of cross-section of wire will become A = ?r² = ?(d2)² = ?(2d2)² = 4A Now R ? 1A, so the resistance will decrease by four times or new resistance will be R = R4 = 1004 = 25?

##### Question 22. The resistance of a wire of 0.01 cm radius is 10 ?. If the resistivity of the material of the wire is 50 × 10-8 ohm meter, find the length of the wire?Year of Question :(2014)

Answer: Here, r = 0.01 cm = 10-4 m, ? = 50 × 10-8 ? m and R = 10 ? As, R = ?lA or l = RA?=R?(?r2) so l = 1050×10?83.14×(10?4)2 = 0.628 m = 62.8 cm

##### Question 23. A wire has a resistance of 16 ?. It is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire. What is the percentage change in its resistance?Year of Question :(2013)

Answer: When wire is melted, its volume remains same, so, V = V or Al = Al Here, l = l2 Therefore, A = 2 A Resistance, R = ?lA = 16 ? Now, R = ?lA=?(l/2)2A=14?lA So, R = R4 = 164 = 4 ? Percentage change in resistance, = (R?RR)×100=(16?416) × 100 = 75%

##### Question 24. If the radius of a current carrying conductor is halved, how does current through it change?Year of Question :(2014)

Answer: If the radius of conductor is halved, the area of cross-section reduced to (14) of its previous value. Since, R ? 1A, resistance will become four times From Ohms law, V = IR For given V, I ? 1R So, current will reduce to one-fourth of its previous value

##### Question 25. 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 cm length of wire of cross sectional area 0.01 square mm and of resistivity 5 × 10-8 ? m.?Year of Question :(2014)

Answer: Resistance is the property of a conductor to resist the flow of charges through it. Factors affecting resistance of a conductor: Rheostat is the device which is often used to change the resistance without changing the voltage source in an electric circuit. We are given, length of wire, l = 50 cm = 50 × 10-2 m cross-sectional area, A = 0.01 mm² = 0.01 × 10-6 m² and resistivity, ? = 5 x 10-8 ? m. As, resistance, R = ?lA ? R = (5×10?8×50×10?20.01×10?6) ? = 2.5 ?

##### Question 26. If a person has five resistors each of value 15 ?, then the maximum resistance he can obtain by connecting them is?Year of Question :(2012)
• (a) 1 ?
• (b) 5 ?
• (c) 10 ?
• (d) 25 ? (2020)

• (a) The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus, Rs = 15+15+15+15+15 1 ?
##### Question 27. The maximum resistance which can be made using four resistors each of 2 ? is?Year of Question :(2012)
• (a) 2 ?
• (b) 4 ?
• (c) 8 ?
• (d) 16 ? (2020)

• (c) : A group of resistors can produce maximum resistance when they all are connected in series. ? Rs = 2 ? + 2 ? + 2 ? + 2 ? = 8 ?
##### Question 28. The maximum resistance which can be made using four resistors each of resistance 12 ? is?Year of Question :(2012)
• (a) 2 ?
• (b) 1 ?
• (c) 2.5 ?
• (d) 8 ? (2020)

• (a) The maximum resistance can be produced from a group of resistors by connecting them in series. Thus, Rs = 12 ? + H 12 ? + 12 ? + 12 ? = 2 ?
##### Question 29. Three resistors of 10 ?, 15 ? and 5 ? are connected in parallel. Find their equivalent resistance.?Year of Question :(2014)

Answer: Here, R1 = 10 ?, R2 =15 ?, R3 = 5 ?. In parallel combination, equivalent resistance, (Req) is given by

##### Question 30. List the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series?Year of Question :(2013)

• (a) When a number of electrical devices are connected in parallel, each device gets the same potential difference as provided by the battery and it keeps on working even if other devices fail. This is not so in case the devices are connected in series because when one device fails, the circuit is broken and all devices stop working
• (b) Parallel circuit is helpful when each device has different resistance and requires different current for its operation as in this case the current divides itself through different devices. This is not so in series circuit where same current flows through all the devices, irrespective of their resistances
##### Question 31. Show how would you join three resistors, each of resistance 9 ? so that the equivalent resistance of the combination is (i) 13.5 ?, (ii) 6 ? ?Year of Question :(2018)

• (i) The resistance of the series combination is higher than each of the resistances. A parallel combination of two 9 ? resistors is equivalent to 4.5 ?. We can obtain 13.5 ? by coupling 4.5 ? and 9 ? in series. So, to obtain 13.5 ?, the combination is as shown in figure (a).
• (ii) To obtain a equivalent resistance of 6 ?, we have to connect two 9 ? resistors in series and then connect the third 9 ? resistor in parallel to the series combination as shown in the figure (b).
##### Question 32. Three resistors of 3 ? each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery?Year of Question :(2017)

Answer: As given in circuit diagram, two 3 ? resistors are connected in series to form R1; so R1 = 3 ? + 3 ? = 6 ? And, R1 and R2 are in parallel combination, Hence, equivalent resistance of circuit (Req) given by Req = 2 ? Using Ohms law, V = IR We get, 3 V = I × 2 ? or I = 32 A = 1.5 A Current drawn from the battery is 1.5 A

##### Question 33. Two identical resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in two cases?Year of Question :(2012)

Answer: Let resistance of each resistor be R. For series combination, Rs = R1 + R2 So, Rs = R + R = 2R For parallel combination,

##### Question 34.
• (a) A 6 ? resistance wire is doubled on itself. Calculate the new resistance of the wire
• (b) Three 2 ? resistors A, B and C are connected in such a way that the total resistance of the combination is 3 ?. Show the arrangement of the three resistors and justify your answer. (2020)

• (a) Given resistance of wire, R = 6 ? Let l be the length of the wire and A be its area of cross-section. Then R = ?lA = 6 ? Now when the length is doubled, l = 2l and A = A2 ? R = ?(2l)A/2=4?lA = 4 × 6 ? = 24 ?
• (b) Given the total resistance of the combination = 3 ? In order to get a total resistance of 3 ?, the three resistors has to be connected as shown. Such that, 1RP=12+12 = 1 ? Rp = 1 ? and Rs = 2 ? + 1 ? = 3 ?
##### Question 35. Draw a schematic diagram of a circuit consisting of a battery of 3 cells of 2 V each, a combination of three resistors of 10 ?, 20 ? and 30 ? connected in parallel, a plug key and an ammeter, all connected in series. Use this circuit to find the value of the following ?Year of Question :(2012)
• (a) Current through each resistor
• (b) Total current in the circuit
• (c) Total effective resistance of the circuit. (2020)

Answer: The circuit diagram is as shown below

• (a) Given, voltage of the battery = 2V + 2V + 2V = 6 V Current through 10 ? resistance, I10 = VR=610 = 0.6 A Current through 20 ? resistance, I20 = VR=620 = 0.3 A Current through 30 ? resistance, I30 = VR=630 = 0.2 A
• (b) Total current in the circuit, 1= I10 + I20 + I30 = 0.6 + 0.3 + 0.2 = 1.1 A (c) Total resistance of the circuit, 1RP=110+120+130=1160
##### Question 36.
• (a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances
• (b) In an electric circuit two resistors of 12 ? each are joined in parallel to a 6 V battery. Find the current drawn from the battery. (Delhi 2019)

• (a) Resistors in parallel : When resistors are connected in parallel.
• (i) The potential difference across their ends is the same
• (ii) The sum of current through them is the current drawn from the source of energy or cell. I = I1 + I1 + I3 or VRP=VR1+VR2+VR3
• (iii) The equivalent resistance is given by, 1RP=1R1+1R2+1R3 Hence equivalent resistance in parallel combination is equal to the sum of reciprocals of the individual resistances
##### Question 37. For the series combination of three resistors current in each resistor, establish the relation R = R1 + R2 + R3 where the symbols have their usual meanings. Calculate the equivalent resistance of the combination of three resistors of 6 ?, 9 ? and 18 ? joined in parallel?Year of Question :(2017)

Answer: Given figure shows the series combination of three resistors R1, R2 and R3 connected across a voltage source of potential difference V. Let current I is flowing through the circuit. V1, V2 and V3 are the potential differences across resistors R1, R2 and R3 respectively. Since, the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors. ? v = v1 + v2 + v3 .(i) In series current through each resistor is same. Applying the Ohms law, V1 = IR1, V2 = IR2 and V3 = IR1 ....(ii) If Rs is the equivalent resistance of the circuit, then V = IRs .(iii) From eqns. (i), (ii) and (iii), we can write IRs = IR1 + IR2 + IR3 or Rs = R1 + R2 + R3 We can conclude that when several resistors are joined in series, the resistance of the combination Rs equals the sum of their individual resistances, R1, R2 and R3 Given : R1 = 6 ?, R2 = 9 ?, R3 = 18 ? are connected in parallel. Equivalent resistance, Req, is given by or Req = 3 ?

##### Question 38. State ohms law. Represent it graphically. In the given circuit diagram calculate?Year of Question :(2012)
• (i) the total effective resistance of the circuit
• (ii) the current through each resistor

Answer: Graphical representation of Ohms law For the given circuit R1 = 3 ?, R2 = 4 ?., R3 = 6 ? and V = 6V

• (i) Total effective resistance of the circuit, Req is given by 1Req=1R1+1R2+1R3=13+14+16=912 or Req = 129 ? = 43 ? = 1.33 ?
• (ii) Since, potential difference across each resistor connected in parallel is same. So, V1 = V2 = V3 = 6 V Applying Ohms law, V1 = I1R1 or I1 = V1R1 or I3 = 63 A = 2A Similarly, I2 = 6A4 = 1.5 A and I3 = 66 A = 1 A
##### Question 40. 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?Year of Question :(2014)

Answer: Change the positions of ammeter and note the reading of ammeter each time. You will find that all the reading obtained are same. So, the value of the current in the ammeter is the same, independent of its position in the electric circuit. It means that in this circuit (series combination) the current is the same in every part of the circuit

Precautions:
• (i) All the connections are neat and tight
• (ii) Ammeter is connected with the proper polarity, i.e., positive terminal of the ammeter should go to positive terminal and negative terminal of ammeter to the negative terminal of the battery or cell used