{"id":13886,"date":"2024-03-21T10:16:55","date_gmt":"2024-03-21T04:46:55","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=13886"},"modified":"2024-03-22T17:34:43","modified_gmt":"2024-03-22T12:04:43","slug":"ts-10th-class-physical-science-important-questions-chapter-9","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-physical-science-important-questions-chapter-9\/","title":{"rendered":"TS 10th Class Physical Science Important Questions Chapter 9 Electric Current"},"content":{"rendered":"

These TS 10th Class Physics Chapter Wise Important Questions<\/a> Chapter 9 Electric Current will help the students to improve their time and approach.<\/p>\n

TS 10th Class Physical Science Important Questions Chapter 9 Electric Current<\/h2>\n

1 Mark Questions<\/span><\/p>\n

Question 1.
\nGive reason, why metals conduct electric charge easily?
\nAnswer:
\nBecause metals possess a large number of free electrons.<\/p>\n

Question 2.
\nHow is an ampere related to a coulomb? (ASL)
\nAnswer:
\n1 amp = \\(\\frac{1 C}{1 \\mathrm{sec}}\\)<\/p>\n

Question 3.
\nWhat is the name of physical quantity which is equal to V\/I?
\nAnswer:
\nElectrical resistance.<\/p>\n

Question 4.
\nHow is one – 1 ohm related to ampere and volt?
\nAnswer:
\n1 ohm= \\(\\frac{1 \\text { volt }}{1 \\text { ampere }}\\)<\/p>\n

Question 5.
\nWhich material is the best conductor?
\nAnswer:
\nSilver.<\/p>\n

Question 6.
\nResistance of an incandescent filament of a lamp is more than that when it is at the room temperature. Why?
\nAnswer:
\nBecause resistance of metallic wire increases with increase in temperature.<\/p>\n

Question 7.
\nWhat is the shape of V – I graph for a metallic wire?
\nAnswer:
\nA straight line passing through origin.
\n\"TS<\/p>\n

Question 8.
\nWhen resistances are connected in series which physical quantity remains unchanged?
\nAnswer:
\nCurrent.<\/p>\n

\"TS<\/p>\n

Question 9.
\nWhen resistances are connected In parallel which physical quantity remains unchanged?
\nAnswer:
\nVoltage.<\/p>\n

Question 10.
\nThe length of a wire is doubled and its cross-sectional area Is also doubled. What is the change \u00a1n its resistivity?
\nAnswer:
\nThere Is no change In resistivity. When the length of the wire is doubled, its resistivity also double. But when the cross-sectFonal areas is doubled, Its resistivity becomes half of the double. So there is no charge.<\/p>\n

Question 11.
\nWhat happens to resistance when lengths of a conductor is doubled without affecting the thickness of conductor?
\nAnswer:
\nResistance is doubled because R\u221d l.
\n\\(\\frac{R_1}{R_2}=\\frac{l_1}{l_2} \\Rightarrow \\frac{R}{R_2}=\\frac{l}{2 l}, \\frac{R}{R_2}=\\frac{1}{2} \\Rightarrow R_2=2 R\\)<\/p>\n

Question 12.
\nA battery of 6V is applied across a resistance of 15c. Find the Current flowing through the circuit.
\nAnswer:
\nCurrent I = \\(\\frac{V}{R}\\), I = \\(\\frac{6}{15} \\) = 0.4 amp<\/p>\n

Question 13.
\nHow is power related to current and voltage?
\nAnswer:
\nPower (P) = Potential difference (V) x Current (I).<\/p>\n

Question 14.
\nHow can we measure potential difference or emf?
\nAnswer:
\nWith the help of a voltmeter, we measure potential difference or emf.<\/p>\n

Question 15.
\nWhat is a conductor of electricity?
\nAnswer:
\nThe material which transfers energy from battery (source) to the bulb is called a conductor A conductor possesses large number of free electrons. Eq: All metals.<\/p>\n

\"TS<\/p>\n

Question 16.
\nWhat is a non-conductor?
\nAnswer:
\nThe material which cannot transfer energy from battery (source) to the bulb is called a non-conductor Electrons in a non-conductor are not free to move.<\/p>\n

Question 17.
\nWhich instrument is used to measure electric current? (AS1)
\nAnswer:
\nAn ammeter is used to measure electric current in a circuit. It is always connected In series to the circuit.<\/p>\n

Question 18.
\nDefine potential difference and give an expression to It.
\nAnswer:
\nElectric potential difference between two points in an electric circuit is the work done to move a unit positive charge from one point to another.
\nPotential difference v = \\(\\frac{W}{q}\\)
\nThe S.I. unit of potential difference is \u2018Volt\u2019 and denoted by \u2018v\u2019<\/p>\n

Question 19.
\nState Ohm\u2019s law.
\nAnswer:
\nOhm\u2019s law: The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature.
\nV \u221d I \u21d2 \\(\\frac{\\mathrm{V}}{\\mathrm{I}} \\) = R<\/p>\n

Question 20.
\nWhat are the limitations of Ohm\u2019s law?
\nAnswer:
\nLimitations of Ohm\u2019s law:<\/p>\n

    \n
  1. Ohm\u2019s law is valid for metal conductors, provided the temperature and other physical conditions remain constant,<\/li>\n
  2. Ohm\u2019s law is not applicable to gaseous conductors.<\/li>\n
  3. Ohm\u2019s law Is also not applicable to semiconductors such as Germanium and Silicon.<\/li>\n<\/ol>\n

    Question 21.
    \nWhat is a resistor?
    \nAnswer:
    \nThe material which offers resistance to the motion of electrons is called resistor.<\/p>\n

    Question 22.
    \nWhat are the uses of semiconductors?
    \nAnswer:
    \nSemiconductors are used to make diodes, transistors, and integrated circuits (ICs). Ic\u2019s are used in all sorts of electronic devices, including computers, TV., mobile phones…. etc.<\/p>\n

    Question 23.
    \nDraw the electric circuit with the help of a Battery, Voltmeter, Ammeter, Resistance and connecting wires.
    \nAnswer:
    \n\"TS<\/p>\n

    Question 24.
    \nWhat do you mean by one \u2018unit\u2019 in household consumption of electrical energy?
    \nAnswer:
    \nOne unit in household consumption of electrical energy is equal to 1 KWH (Kilo Watt Hour)
    \n1 KWH 1000 W x 1 Hour
    \n(1000)W x (60 x 60) sec
    \n1000 J\/s x 3600 sec = 36 x 10 Joules.<\/p>\n

    Question 25.
    \nWhen do you say that two or more resistors are connected in series?
    \nAnswer:
    \nTwo or more resistors are said to be connected In series If the same current flows through them.<\/p>\n

    Question 26.
    \nWhen do you say that two or more resistors are connected in parallel?
    \nAnswer:
    \nTwo or more resistors are said to be connected In parallel if the same potential difference exists across them.<\/p>\n

    Question 27.
    \nWhat is lattice?
    \nAnswer:
    \nAccording to Drude and Lorentz, conductors like metals contain a large number of free electrons, while the positive ions are fixed in their locations. The arrangement of the positive ions Is called lattice.<\/p>\n

    Question 28.
    \nWhy do electrons move in specified direction when the conductor is connected to a battery?
    \nAnswer:
    \nWhen the ends of the conductor are connected to the terminals of a battery, a uniform electrical field is set up throughout the conductor. This field makes the electrons move in a specified direction.<\/p>\n

    Question 29.
    \nWhich instrument is used to measure potential difference or CML?
    \nAnswer:
    \nA volt meter Is used to measure potential difference or emf across an electric device like battery. It must be connected parallel to the electric device.<\/p>\n

    Question 30.
    \nWhat is drift speed of electrons?
    \nAnswer:
    \nThe electrons in a conductor move with constant average speed, known as drift speed or drift velocity.<\/p>\n

    Question 31.
    \nIs the voltmeter connected in series or parallel In circuit? Why?
    \nAnswer:
    \nVoltmeter should be connected parallel in the circuit to measure the potential difference between two points of conductor.<\/p>\n

    \"TS<\/p>\n

    Question 32.
    \nState the use of Ammeter? How to connect the Ammeter in electric circuit?
    \nAnswer:
    \nAmmeter is used to measure electric current In a circuit. It should be connected in series in a circuit.<\/p>\n

    Question 33.
    \nThe home appliances like Fridge, T.V, Computer are connected In series or parallel? Why?
    \nAnswer:
    \nThey are connected in parallel because If any one device is damaged rest will work as usual because the circuit does not break.<\/p>\n

    Question 34.
    \nWhy are copper wires used as connecting wires?
    \nAnswer:
    \nCopper is a good conductor of electricity so copper wires are used as connecting wires.<\/p>\n

    Question 35.
    \nName two special characteristics of fuse wire.
    \nAnswer:
    \nHigh resistivity and low melting point.<\/p>\n

    Question 36.
    \nName two special characteristics of heating coil.
    \nAnswer:
    \nHigh resistivity and high melting point.<\/p>\n

    Question 37.
    \nWhat is Resistance? What are the SI Units of Resistance?
    \nAnswer:
    \nResistance of a conductor is defined as the obstruction to the motion of the electrons In a conductor. It\u2019s S.I unit is Ohm.<\/p>\n

    Question 38.
    \nWhat happens if we use a fuse made up of same wire which is used to make the electric circuit?
    \nAnswer:
    \nThe fuse made of same wire cannot gets heated up and melts due its low resistance when excess current is drawn from the mains. Due to this, the electrical appliances In the house wiii be damaged.<\/p>\n

    2 Marks Questions<\/span><\/p>\n

    Question 1.
    \nGive reasons for using lead in making fuses.
    \nAnswer:<\/p>\n

      \n
    • Let’s used in making fuses because it has low melting point E.,, resistivity.<\/li>\n
    • If the current in the lead wire exceeds certain value the wire will heat up and melt, so the circuit in the households, opened and ail the electric devices are saved.<\/li>\n<\/ul>\n

      Question 2.
      \nDefine electric current and give an expression to It.
      \nAnswer:
      \nElectric current is defined as the amount of charge crossing any cross-section of the conductor in one second.
      \nLet Q be the charge crossing any cross-section of the conductor In time \u2018t\u2019.
      \n\u2234 Electric current = \\( \\)
      \nI = \\(\\frac{Q}{t} \\)
      \nThe SI unit of current \u00a1s \u2018ampere\u2019 denoted by \u2018A\u2019.<\/p>\n

      Question 3.
      \nWhat is drift speed?
      \nAnswer:
      \nElectrons in the conductor move w,th a constant average speed, which is known as drift speed or drift velocity.
      \nDrift velocity V = \\(\\frac{1}{\\mathrm{nqA}} \\)
      \nWhere, I = Electric current
      \nn = number of charges
      \nq = magnitude of electric charge
      \nA = Area of cross-section of the conductor<\/p>\n

      Question 4.
      \nDefine emf.
      \nAnswer:
      \nElectromotive force or emf is defined as the work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.
      \nemf \u03b5 = \\(\\frac{W}{q}=\\frac{F_e d}{q} \\)
      \nWhere Fe<\/sub> is electric force
      \nd is the distance between the terminals
      \nq is the charge
      \nThe S.l unit of emf is \u2018Volt\u2019.<\/p>\n

      Question 5.
      \nWhat are the factors affecting the resistance of a material?
      \nAnswer:<\/p>\n

        \n
      1. The value of resistance of a conductor depends on temperature for constant potential difference.<\/li>\n
      2. Resistance of a conductor depends on the material of the conductor.<\/li>\n
      3. Resistance of a conductor is directly proportional to its length i.e., R \u221d l<\/li>\n
      4. Resistance of a conductor is inversely proportional to the area of cross-section of the material. i.e., R\u221d \\(\\frac{l}{A} \\)<\/li>\n<\/ol>\n

        Question 6.
        \nFind the resistance of a bulb, on which 60W and 1.20 V is marked.
        \nAnswer:
        \n60w, V= 120 V
        \nWe know P=VI \u21d2 \\(\\frac{\\mathrm{V}^2}{\\mathrm{R}} \\)
        \n\u21d2 R = \\(\\frac{\\mathrm{V}^2}{\\mathrm{R}}=\\frac{120 \\times 120}{60} \\) = 240 \u2126<\/p>\n

        \"TS<\/p>\n

        Question 7.
        \nWrite any two differences between ohmic and non-ohmic conductors.
        \nAnswer:<\/p>\n\n\n\n\n\n\n
        Ohmic conductors<\/td>\nNon-ohmic conductors<\/td>\n<\/tr>\n
        1) Ohmic conductors follow the ohm\u2019s law<\/td>\n1) Non-ohmic conductors do not follow the ohm\u2019s law<\/td>\n<\/tr>\n
        2) Ohmic conductors are electric conductors.<\/td>\n2) Non-ohmic conductors are semiconductors.<\/td>\n<\/tr>\n
        3) V-I graph of ohmic conductors is a straight line is a curve.<\/td>\n\u00a03) V-I graph of non-ohmic conductors<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Question 8.
        \nTwo wires have equal area of cross-section. One Is copper and other Is Aluminium have the same resistance. Find which one Is longer.
        \nAnswer:
        \nSuppose the resistance of copper and aluminum, wires are R1<\/sub> and R2<\/sub>, Suppose their area of cross section is A
        \nThe resistivity of copper (P1<\/sub>) = 1.68 x 10-8<\/sup>
        \nThe resistivity of aluminum. (P2<\/sub>) = 2.82 x 10-8<\/sup>
        \nGiven that R1<\/sub> = R2<\/sub>
        \n\"TS
        \nLength of copper wire 1.678 times more than length of aluminum wire.<\/p>\n

        Question 9.
        \nDefine Ohmic and non-ohmic conductors and give two examples each of them.
        \nAnswer:
        \nOhmic conductors: The conductors which does not obey Ohm\u2019s law are called non-ohmic conductors. e.g.: Semiconductors, Electrolytes.<\/p>\n

        Question 10.
        \nThree equal resistances are connected in series then in parallel. What will be the ratio of their resultant resistances?
        \nAnswer:
        \nSuppose the resistance of equal resistors is \u2018R\u2019. Suppose they are connected in series.
        \nThen their equivalent resistance R1<\/sup> = R + R + R = 3R
        \nIf they are connected in parallel their equivalent resistance.
        \nFrom \\(\\frac{1}{R^{11}}=\\frac{1}{R}+\\frac{1}{R}+\\frac{1}{R}=\\frac{3}{R} \\)
        \n\\(\\frac{1}{R^{11}}=\\frac{3}{R} \\Rightarrow R^{11}=\\frac{R}{3} \\)
        \nRatio of resultant resistances = R1<\/sup> : R11<\/sup>= \\(3 \\mathrm{R}: \\frac{\\mathrm{R}}{3}\\) = 9 : 1<\/p>\n

        Question 11.
        \nWrite differences between overloading and short-circuiting.
        \nAnswer:
        \nCurrently chooses the path which has least resistance. So when electricity travels along a wrong route because of damaged wires or a fault in the connections, it leads to burn. This is known as short circuit.
        \nWhen so many electrical appliances are connected to the same electrical main point then maximum current can be drawn from the mains which causes overheating and may cause a fire which is called \u2018overloading\u2019.<\/p>\n

        4 Marks Questions<\/span><\/p>\n

        Question 1.
        \nA circuit Is shown in the picture. 200 c \u00b0 The current passing through A Is I
        \na) What is the potential difference between A and B?
        \nb) What is the equivalent resistance between A and B?
        \nc) What amount of current is flown through C and D?
        \n\"TS
        \nAnswer:
        \na) According to Kirchhoff’s loop law the algebraic sum of increase and decrease In p.d across various components of the circuit In a closed circuit loop must be zero. So the p.d across CD is zero because it is a closed loop.
        \nb) Here 20\u2126, 5\u2126 are parallel to each other and resultants are In series to each other. Resultant resistance of 20\u2126 and 5\u2126.
        \nWhen they are connected in parallel is \\(\\frac{1}{\\mathrm{R}^{\\prime}}=\\frac{1}{20}+\\frac{1}{5}=\\frac{1+4}{20}=\\frac{5}{20}=\\frac{1}{4}\\)
        \n\\(\\frac{1}{\\mathrm{R}^{\\prime}}=\\frac{1}{4} \\Rightarrow \\mathrm{R}=4 \\Omega \\)
        \n(c) VA<\/sub> – VD<\/sub> = VA<\/sub> – VC<\/sub>, So
        \n5I – 5i = 20i \u21d2 20i+5i \u21d2 5I = 25i \u21d2 i = \\(\\frac{5 \\mathrm{I}}{25}=\\frac{1}{5}\\)
        \nCurrent through CD = I – 2i = I – \\(\\frac{2 I}{5}=\\frac{3}{5}\\) I Amp<\/p>\n

        Question 2.
        \nObserve the picture. The potential values at A, B, C are 70V, 0V, 10V
        \na) What is the potential at D?
        \nb) Find the ratio of the flow of B current in AD, DB, DC.
        \n\"TS
        \nAnswer:
        \na) By following Ohm\u2019s law. p.d Is (V) = iR
        \nIn the given circuit we are applying junction laws.
        \n\u2018D\u2019 works as a junction so, i = i1<\/sub> +i2<\/sub>
        \nLet p.d at D is V0<\/sub>.
        \nWe know that i = \\(\\frac{V}{R} \\)
        \n\"TS
        \ni = \\(\\frac{V}{R}=\\frac{V_A-V_0}{R}=\\frac{70-V_0}{10}\\)
        \n\"TS
        \nBy the law i = i1<\/sub> +i2<\/sub>
        \n\"TS
        \n\u2234 Potential at D is = V0<\/sub> = 40 V<\/p>\n

        b)Flow of current in AD is i= \\(\\frac{\\mathrm{V}}{\\mathrm{R}}=\\frac{\\mathrm{V}_{\\mathrm{A}}-\\mathrm{V}_0}{\\mathrm{R}}=\\frac{70-40}{10}=\\frac{30}{10}=3\\)
        \n(\u2234 Here VA<\/sub> =70, V0<\/sub>=40, R=10)<\/p>\n

        Flow of current in DB is i1<\/sub> = \\(\\frac{V_0-V_B}{R_1}=\\frac{40-0}{20}=\\frac{40}{20}=2 \\)
        \n(\u2234 Here V0<\/sub>=40, VB<\/sub>=0, R1<\/sub> = 20)<\/p>\n

        Flow of current in DC is i2<\/sub> = \\(\\frac{\\mathrm{V}_0-\\mathrm{V}_{\\mathrm{C}}}{\\mathrm{R}_2}=\\frac{40-10}{30}=\\frac{30}{30}=1 \\)
        \n(\u2234 Here V0<\/sub>=40, VC<\/sub>=0, R2<\/sub> = 30)
        \nThe ratio of the flow of current In AD, DB, and Dc is 3:2:1.s<\/p>\n

        \"TS<\/p>\n

        Question 3.
        \nIn a circuit, 60V battery, three resistances R1<\/sub> = 10\u03a9, R2<\/sub> = 20\u03a9 and R3<\/sub> = X \u03a9 are connected In series. If 1-ampere current flows in the circuit, find the resistance in R3<\/sub> by using Kirchhoff\u2019s loop law.
        \nAnswer:
        \nAccording to Loop law,
        \n60-101- 201-XI=0
        \nsubstituting I = 1 Amp, In the above equation, R
        \n60 – 10 – 20 – x – 0
        \nx = 30 \u03a9
        \n\u2234 R3<\/sub> =30 \u03a9
        \n\"TS<\/p>\n

        Another method:
        \nR1<\/sub> = 10 \u03a9, R2<\/sub> = 20 \u03a9, R3<\/sub> = X \u03a9
        \nAs they are connected in series, the resultant
        \nResistantR = R1<\/sub> + R2<\/sub>+ R3<\/sub>
        \n= 10 \u03a9 + 20 \u03a9 +X \u03a9
        \n= 30 \u03a9 + X \u03a9
        \nI = 1 Amp, V = 60v<\/p>\n

        According to Ohm\u2019s law,
        \nV=IR
        \n60 V = I x (30 + X\u03a9)
        \n60 V = 30 +X\u03a9
        \nx = 60- 30
        \nX = 30
        \n\u2234 R3<\/sub> = 30\u03a9<\/p>\n

        Question 4.
        \nA circuit is made with a copper wire as shown in the diagram. We know that conductor\u2019s resistance is directly proportional to its length. Calculate the equivalent resistance between points 1 and 2.
        \n\"TS
        \nAnswer:
        \nLet the resistance of the wire be \u2018R\u2019 and length of the wire be l.
        \nLet the length be \u2018l\u2019
        \nIn a square diagonal is \\(\\sqrt{2}\\) times its length = \\(\\sqrt{2}\\) l
        \nResistance towards diagonal is \\(\\sqrt{2}\\)R.
        \n\"TS
        \nThe circuit diagrams for the given arrangement is along PTR and QTS Is ineffective as no current flow through it.
        \nPQ and PS are in series so effective resistance are R1<\/sub>+ R2<\/sub> = R + R = 2R.
        \nQR and SR are In series so effective resistance are R1<\/sub> + R3<\/sub> = R + R = 2R.
        \nRedrawn of the circuit again as resultant resistance between the points 1 and 2 is
        \n\"TS<\/p>\n

        Question 5.
        \nDerive an expression to find drift velocity of electrons.
        \nAnswer:
        \n1. Consider a conductor with cross-sectional area A. Assume that the two ends of the conductor are connected to a battery to make the current flow through it.
        \n\"TS
        \n2. Let \u2018Vd<\/sub>\u2019 be the drift speed of the charges and \u2018n\u2019 be the number of charges present in the conductor in an unit volume.
        \n3. The distance covered by cacti charge in one second is \u2018Vd<\/sub>\u2019
        \n4. Then the volume of the conductor for this distance. A\u2019Vd<\/sub>‘
        \n5. The number of charges contained in that volume = n. A\u2019Vd<\/sub>\u2019
        \n6. Let q be the charge of each carrier.
        \n7. Then the total charge crossing the cross-sectional area at position D in one-second in q Avd<\/sub>\u2019.
        \nThis is equal to electric current.
        \nElectric current I = n q Avd<\/sub>.
        \nDrift velocity (Vd<\/sub>) = \\(\\frac{1}{n q A} \\)<\/p>\n

        Question 6.
        \nShow that the semiconductors do not obey Ohm\u2019s law.
        \nAnswer:
        \n1. Connect a circuit as shown n the figure.
        \n\"TS
        \n2. Close the circuit and note the readings of ammeter (I) and voltmeter (V) tri the following table.<\/p>\n\n\n\n\n\n\n
        Potential difference (V)<\/td>\nCurrent (I)<\/td>\nV\/I<\/td>\n<\/tr>\n
        <\/td>\n<\/td>\n<\/td>\n<\/tr>\n
        <\/td>\n<\/td>\n<\/td>\n<\/tr>\n
        <\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        3. Now connect two cells Instead of one cell in circuit.
        \n4. Now note the respective readings of ammeter and voltmeter.
        \n5. Repeat the same for 3, 4, and 5 cells.
        \n6. Find the ratio of every time.
        \n7. We find that the ratio Is not constant.
        \n8. Draw a graph between V and I. You will get a graph as shown below.
        \n9. This shows the semiconductors (LED is a semiconductor) do not obey Ohm\u2019s law.
        \n\"TS<\/p>\n

        Question 7.
        \nDeduce an expression to measure electric power.
        \nAnswer:
        \n1. Consider that a charge (Q) Coulomb passes through a point A, moves to print B in the time interval \u2018t\u2019 seconds.
        \n\"TS
        \n2. Let V be the potential difference between the points A and B.
        \n3. The work done by electric field in time \u2018t\u2019 is given by W = QV …………………… (1)
        \n4. The work is equal to the energy lost by the charge while passing through the conductor for time T.
        \n5. Energy lost by the conductor in 1 sec = W\/t
        \nFrom (1) W\/t = QV\/t ………………….. (2)
        \nWe know, Q\/t = I and W\/t = P (power)
        \nThen(2) \u21d2P =VI ………………………….. (3)
        \nThis equation can be used to calculate power consumption by any electric device that is connected in a circuit.
        \nFrom Ohm\u2019s law V = I R
        \n(3) \u21d2 P = I2<\/sup> R. (or) V2<\/sup>\/R ……………………. (4)
        \n6. To know the power that can be extracted from a battery or any source can be calculated by P = \u03b5J.
        \nWhose \u03b5 the is the emf of the battery or source.<\/p>\n

        \"TS<\/p>\n

        Question 8.
        \nFind the resultant resistance for the following given arrangement. Find the current, when this arrangement Is connected with 9V battery.
        \n\"TS
        \nAnswer:
        \n\"TS
        \nFrom the above circuit, all the resistors are in parallel combination
        \n\u2234 \\(\\frac{1}{R}=\\frac{1}{R_1}+\\frac{1}{R_2}+\\frac{1}{R_3}\\)
        \nHence R1<\/sub>= R2<\/sub>=R3<\/sub> = 3W
        \n\u2234 \\(\\frac{1}{\\mathrm{R}}=\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{3} \\)
        \n\\(\\frac{1}{R}=3\\left(\\frac{1}{3}\\right)=1\\)
        \n\u2234 R = 1 W
        \nCurrent given V = 9V
        \n\u2234 i =\\(\\frac{\\mathrm{V}}{\\mathrm{R}}=\\frac{9}{1} \\) = 9A<\/p>\n

        Question 9.
        \nObserve the circuit given. R1<\/sub> = R2<\/sub> = R3<\/sub> = 200 \u2126.
        \nThen find out the electromotive force E of battery.
        \nAnswer:
        \nVoltmeter reading V= 100 V, Voltmeter resistance = 1000
        \nCurrent i1<\/sub> = \\(\\frac{V}{R}=\\frac{100}{1000} \\) = 0.1 Amperes
        \n\"TS
        \nConsidering ABDA, applying Kirchoff\u2019s voltage law, R,
        \n1000i1<\/sub> + 200i = 2E
        \n1000 x 0.1 + 200i = 2E
        \n100+200i =2E …………………………. (1)<\/p>\n

        Considering BCDB, applying Kirchofr\u2019s voltage law,
        \n(i-i1<\/sub>)200 + 200( i-i1<\/sub>)-1000 x i1<\/sub>=0
        \n\"TS<\/p>\n

        Question 10.
        \nThe electric circuit is shown in the figure. Find out the equivalent resistance between A and B?
        \nAnswer:
        \nThe first three resistors are in parallel.
        \n\"TS
        \n\\(\\frac{1}{R_{e f f}}=\\frac{1}{R}+\\frac{1}{R}+\\frac{1}{R}=\\frac{3}{R}\\)
        \nRef<\/sub> = \\(\\frac{R}{3} \\)
        \nAgain \\(\\frac{R}{3} \\) and R are connected in series.
        \n\"TS
        \nRef<\/sub> = R1<\/sub>+R2<\/sub> = \\(\\frac{R}{3}+R=\\frac{4 R}{3}\\)
        \n\"TS<\/p>\n

        Question 11.
        \nSudhakar has taken a substance in the form of wire. He applied different voltages to the wire and measured electrical currents. For this he used Ammeter and Voltmeter. He tabulated five measurements. Then plotted a graph as shown in the figure. In the graph he measured voltages in volts (V) and current (I) in Amperes.
        \nAnswer the following
        \na) WhattypeofmaterlaiSudhakarselected for his experiment?
        \nb) What is the resistance of the substance?
        \nc) If potential difference is 20 Vat the ends of wire. How much electrical power is utilized by wire?
        \nd) What is the law associated with the above graph?
        \n\"TS
        \nAnswer:
        \na) The graph is a straight line passing through origin. it is in the form of y = mx i.e., I mv. Here \u2018m\u2019 is slope of the graph.
        \nHere m = \\(\\frac{1}{R}=\\frac{1}{\\operatorname{Resistance} \\)
        \n\u2234 I = \\(\\left(\\frac{1}{R}\\right) V \\)<\/p>\n

        The substance is Ohmic substance i.e., obeying Ohm\u2019s law. So it Is a metal like iron spoke, (or) Copper, Aluminium etc.<\/p>\n

        b) The resistance can be known from graph is
        \nV = IR; R = \\(\\frac{V}{I}=\\frac{10}{0 \\times 2}=\\frac{100}{2} \\) = 30 \u2126
        \nThe reciprocal of slope of graph gives Resistance.<\/p>\n

        c) The electrical power can be measured by taking the area of graph i.e., area enclosed between the straight line and X – axis
        \nPower (P) = Voltage x Current = V I
        \nArea = Area of tnangle = \\(\\frac{1}{2} \\) x 20 x 0.4 = 4 Watts
        \nd) Ohm\u2019s Law: The potential difference between the ends of a conductor is directly proportional to the electric current passing through It at constant temperature.<\/p>\n

        \"TS<\/p>\n

        Question 12.
        \nYour friend needs 10 ohms resistance. He came to you and asked, but you have 40 ohms resistance.
        \ni) How many resistors your friend will ask you?
        \nii) How the resistors which are taken are connected
        \niii) Show that their effective resistance is lo ohms.
        \nAnswer:
        \ni) A minimum of four resistors are required.
        \nii) They should be connected In parallel.
        \niii) When the resistors are connected in parallel, the equivalent resistance is given
        \n\"TS
        \nR = 10 \u2126<\/p>\n

        Question 13.
        \n12 V battery is connected in a circuit and to this 4\u2126, 12\u2126 resistors are connected in parallel, 3\u2126 resistor is connected in series to this arrangement. Draw the electric circuit from this information and find the current in the circuit.
        \nAnswer:
        \nThe resultant resistance of 4\u2126, 12\u2126, connected in parallel is
        \n\"TS
        \n\"TS
        \nThe total resistance in the circuit is
        \nR = R1<\/sub> +R2<\/sub> = 3 + 3 = 6
        \n\u2234 R = 6\u2126<\/p>\n

        The current in the circuit
        \nI = \\(\\frac{V}{R} \\Rightarrow I=\\frac{12}{6}\\)
        \n\u2234 I = 2A<\/p>\n

        Question 14.
        \nDraw the shape of V – I graph for a conductor and a semiconductor.
        \nAnswer:
        \n\"TS<\/p>\n

        Question 15.
        \nDraw the symbols of the following.
        \ni) Battery
        \nii) Resistance
        \niii) Ammeter
        \niv) Voltmeter
        \nv) Key
        \nvi) Rheostat
        \nAnswer:
        \n\"TS<\/p>\n

        Solved Example<\/span><\/p>\n

        Question 1.
        \nFind electric current drawn (figure) from the battery of emf 12V.
        \nAnswer:
        \nLet I = I1<\/sub>+I2<\/sub> be the current drawn from emf 12V.
        \nFrom the figure.
        \nUsing the loop law, for the loop DABCD,
        \n– 3(I1<\/sub> + I2<\/sub>) + 12 – 2I1<\/sub> – 5 = 0 …………………….. (a)
        \nfor the loop DAFED, .
        \n– 3(I1<\/sub> + I2<\/sub>) + 12 – 4I2<\/sub> = O …………………. (b)
        \nSolving the equation (a) & (b)
        \nWe get I1<\/sub> = 0.5 A and I2<\/sub>= 1.5A
        \nTotal current drawn Is then I = 0.5 + 1.5 = 2A B
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Do You Know<\/span><\/p>\n

        A multimeter is an electric measuring instrument that combines several measurement functions in one unit. A digital multimeter displays the measured value in numerals.
        \nA multi-meter has three parts.
        \n\"TS
        \nDisplay: The display usually has four digits and the ability to display a negative sign.
        \nSelection knob: The selection knob allows the user to set the multimeter to read different functions such as milliamps (mA) of current, voltage (V), and resistance (\u2126).<\/p>\n

        Ports: Multi-meters generally have two ports. One is usually labeled as \u2018COM\u2019 (common or ground port). This is where black test lead is connected. The other is labeled as mAV\u2126 port where the red lead is conventionally plugged in.<\/p>\n

        Warning: Most multimeters can measure AC quantities also, but AC circuits can be dangerous. So measure DC quantities only. (Page 188)<\/p>\n","protected":false},"excerpt":{"rendered":"

        These TS 10th Class Physics Chapter Wise Important Questions Chapter 9 Electric Current will help the students to improve their time and approach. TS 10th Class Physical Science Important Questions Chapter 9 Electric Current 1 Mark Questions Question 1. Give reason, why metals conduct electric charge easily? Answer: Because metals possess a large number of … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13886"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=13886"}],"version-history":[{"count":6,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13886\/revisions"}],"predecessor-version":[{"id":14349,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13886\/revisions\/14349"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=13886"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=13886"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=13886"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}