{"id":35772,"date":"2022-09-30T10:08:46","date_gmt":"2022-09-30T04:38:46","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35772"},"modified":"2022-09-30T10:08:53","modified_gmt":"2022-09-30T04:38:53","slug":"ap-inter-2nd-year-physics-study-material-chapter-10","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-inter-2nd-year-physics-study-material-chapter-10\/","title":{"rendered":"AP Inter 2nd Year Physics Study Material Chapter 10 Alternating Current"},"content":{"rendered":"

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material<\/a> 10th Lesson Alternating Current Textbook Questions and Answers.<\/p>\n

AP Inter 2nd Year Physics Study Material 10th Lesson Alternating Current<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nA transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the secondary if the primary has 10 turns. (A.P. Mar.’19; T.S. Mar. ’16 )
\nAnswer:
\n\\(\\frac{\\mathrm{V}_{\\mathrm{s}}}{\\mathrm{V}_{\\mathrm{p}}}\\) = \\(\\frac{N_s}{N_p}\\)
\nVp<\/sub> = 200V, Vs<\/sub> = 2000V, Np<\/sub> = 10
\nNs<\/sub> = \\(\\frac{\\mathrm{V}_{\\mathrm{s}}}{\\mathrm{V}_{\\mathrm{p}}} \\times \\mathrm{N}_{\\mathrm{p}}\\) = \\(\\frac{2000}{200} \\times 10\\)
\nNs<\/sub> = 100.<\/p>\n

Question 2.
\nWhat type of transformer is used in a 6V bed lamp ?
\nAnswer:
\nStep down transformer is used in 6V bed lamp.<\/p>\n

\"AP<\/p>\n

Question 3.
\nWhat is the phenomenon involved in the working of a transformer ? (T.S. Mar.19; A.P. Mar. 16, Mar. 14)
\nAnswer:
\nTransformer works on the principle of mutual induction.<\/p>\n

Question 4.
\nWhat is transformer ratio ?
\nAnswer:
\nThe ratio of secondary e.m.f to the primary e.m.f. (or) number of turns in secondary to the number of turns in the primary is called the transformer ratio.
\nTransformer ratio = \\(\\frac{V_s}{V_p}\\) = \\(\\frac{N_s}{N_p}\\).<\/p>\n

Question 5.
\nWrite the expression for the reactance of<\/p>\n

    \n
  1. an inductor and<\/li>\n
  2. a capacitor.<\/li>\n<\/ol>\n

    Answer:<\/p>\n

      \n
    1. Inductive reactance (XL<\/sub>) = \u03c9L<\/li>\n
    2. Capacitive reactance (XC<\/sub>) = \\(\\frac{1}{\\omega C}\\)<\/li>\n<\/ol>\n

      Question 6.
      \nWhat is the phase difference between A.C emf and current in the following: Pure resistor, pure inductor and pure capacitor.
      \nAnswer:<\/p>\n

        \n
      1. In pure resistor A.C. e.in.f and current are in phase with each other.<\/li>\n
      2. In pure inductor, current lags behind the e.m.f. by an angle of \\(\\frac{\\pi}{2}\\) (or) 90\u00b0.<\/li>\n
      3. In pure capacitor, current leads the e.m.f by an angle \\(\\frac{\\pi}{2}\\) (or) 90\u00b0.<\/li>\n<\/ol>\n

        Question 7.
        \nDefine power factor. On which factors does power factor depend ?
        \nAnswer:
        \nThe ratio of true power and apparent power (virtual power) in an a.c circuit is called as power factor of the circuit.
        \nPower factor (cos\u03d5) = \\(\\frac{\\mathrm{P}}{\\mathrm{P}_{\\mathrm{rms}}}\\). (\u2234 Prms<\/sub> = Vrms<\/sub>Irms<\/sub>)
        \nPower factor depends on r.m.s voltage, r.m.s current and average power (P).<\/p>\n

        Question 8.
        \nWhat is meant by wattless component of current ?
        \nAnswer:
        \nAverage power (P) = Vrms<\/sub>(Irms<\/sub> sin\u03d5)cos \\(\\frac{\\pi}{2}\\)
        \nThe average power consumed in the circuit due to Irms<\/sub> sin \u03d5) component of current is zero. This component of current is known as wattless current. (Irms<\/sub> sin \u03d5) is the wattless component of current.<\/p>\n

        Question 9.
        \nWhen does a LCR series circuit have minimum impedance?
        \nAnswer:
        \nIn LCR series circuit, Impedance (Z) = \\(\\sqrt{R^2+\\left(\\frac{1}{\\omega C}-\\omega L\\right)^2}\\)
        \nAt a particular frequency, \u03c9L = \\(\\frac{1}{\\omega C}\\)
        \nThe impedance is minimum (Z = R)
        \nThis frequency is called resonant frequency<\/p>\n

        \"AP<\/p>\n

        Question 10.
        \nWhat is the phase difference between voltage and current when the power factor in LCR series circuit is unity ?
        \nAnswer:
        \nIn LCR series circuit power factor (cos \u03d5) = 1
        \n\u2234 Phase difference between voltage and current is zero. (\u03d5 = 0)<\/p>\n

        Short Answer Questions<\/span><\/p>\n

        Question 1.
        \nObtain an expression for the current through an inductor when an AC emf is applied.
        \nAnswer:
        \nCircuit consists of pure inductor of inductance L. Let an ac e.m.f V = Vm<\/sub> sin \u03c9t is applied to it. Let i be the instantaneous current.
        \nBack e.m.f developed across the inductor = -L\\(\\frac{\\mathrm{di}}{\\mathrm{dt}}\\)
        \n\"AP
        \nTotal e.m.f = Vm<\/sub> sin \u03c9t – L\\(\\frac{\\mathrm{di}}{\\mathrm{dt}}\\) ——- (1)
        \nAccording to ohms law this must be equal to iR = 0
        \n\"AP
        \nThis is the expression for the instantaneous current through inductor. Here current lags behind the e.m.f by \\(\\frac{\\pi}{2}\\) radian (or) 90\u00b0.<\/p>\n

        Question 2.
        \nObtain an expression for the current in a capacitor when an AC emf is applied.
        \nAnswer:
        \nCircuit consists of pure capacitor of capacitance C. Let an A.C emf V = Vm<\/sub> sin \u03c9t is applied to it. Let i and q be the instantaneous values of current and charge.
        \nPotential difference across capacitor = –\\(\\frac{\\mathrm{q}}{\\mathrm{C}}\\).
        \nTotal emf = Vm<\/sub> sin \u03c9t –\\(\\frac{\\mathrm{q}}{\\mathrm{C}}\\)
        \n\"AP
        \nAccording to Ohms law this must be equal to iR = 0
        \nVm<\/sub> sin \u03c9t –\\(\\frac{\\mathrm{q}}{\\mathrm{C}}\\) = 0
        \n\"AP
        \ni0<\/sub> is peak value of current. Here the current leads the applied e.m.f. by \\(\\frac{\\pi}{2}\\) or (or) 90\u00b0<\/p>\n

        \"AP<\/p>\n

        Question 3.
        \nState the principle on which a transformer works. Describe the working of a transformer with necessary theory.
        \nAnswer:
        \nTransformer is a device to convert a low alternating current of high voltage into high alternating current of low voltage and vice versa.
        \nPrinciple : It works on the principle of mutual induction between two coils.
        \nWorking : When an alternating emf is applied across the primary coil, the input voltage changes with time. Hence the magnetic flux through the primary also changes with time.
        \nThis changing magnetic flux will be linked with secondary through the core. An emf is induced in the secondary.
        \n\"AP
        \nTheory: Let N1<\/sub> and N2<\/sub> be the number of turns in the primary and secondary. Let VP<\/sub> and Vs be the emf s across the primary and secondary.
        \n\"AP
        \nEfficiency of transformer:
        \nIt is the ratio of output power to the input power.
        \n\"AP<\/p>\n

        Long Answer Questions<\/span><\/p>\n

        Question 1.
        \nObtain on expression for impedance and current in series LCR circuit. Deduce an expresssion for the resonating frequency of an LCR series resonating circuit
        \nAnswer:
        \nCircuit consists of resistor of resistance R, inductor of inductance L and capacitor of capacitance C connected in series with an a.c Voltage V = Vm<\/sub> sin \u03c9t.
        \nLet i be the current and q be the charge at any instant t.
        \n\"AP
        \nBack e.m.f across inductor is -L\\(\\frac{\\mathrm{di}}{\\mathrm{dt}}\\)
        \nand across capacitor is \\(\\frac{-\\mathrm{q}}{\\mathrm{C}}\\)
        \nTotal e.m.f = Vm<\/sub> sin \u03c9t – L\\(\\frac{\\mathrm{di}}{\\mathrm{dt}}\\) – \\(\\frac{\\mathrm{q}}{\\mathrm{C}}\\)
        \nAccording to Ohms law, this must be equal to iR
        \nVm<\/sub> sin\u03c9t – L\\(\\frac{\\mathrm{di}}{\\mathrm{dt}}\\) – \\(\\frac{q}{C}\\) = iR
        \nL\\(\\frac{\\mathrm{di}}{\\mathrm{dt}}\\) + iR + \\(\\frac{q}{C}\\) = Vm<\/sub> sin \u03c9t ——- (1)
        \nThe current i any instant in the circuit is
        \ni = im<\/sub> sin(\u03c9t – \u03d5) if \u03c9L > \\(\\frac{1}{\\omega \\mathrm{C}}\\) which is possible at high frequencies.
        \ni = im<\/sub> sin (\u03c9t + \u03d5) if \u03c9L which is possible at low frequencies.
        \nThe maximum current (im<\/sub>) is given by
        \n\"AP
        \nLet \u03d5 be the phase difference between current and e.m.f
        \n\"AP<\/p>\n

        Resonant frequency(f0<\/sub>):
        \nAt this frequency, the impedance of LCR circuit is minimum and is equal to R. At this frequency current is maximum This frequency is called resonant frequency (f0<\/sub>).
        \n\"AP
        \nIn frequency response curve, at resonant frequency (f0<\/sub>), current is maximum. This series resonant circuit is called acceptor circuit.<\/p>\n

        Problems<\/span><\/p>\n

        Question 1.
        \nAn ideal inductor (no internal resistance for the coil) or 20 mH is connected in series with an AC ammeter to an AC source whose emf is given by e = 20\\(\\sqrt{2}\\) sin(200t + \u03c0\/3)V, where t is in seconds. Find the reading of the ammeter ?
        \nSolution:
        \nGiven that L = 20 mH = 20 \u00d7 10-3<\/sup>H
        \n\"AP<\/p>\n

        Question 2.
        \nThe instantaneous current and instantaneous voltage across a series circuit containing resistance and inductance
        \nare given by i = \\(\\sqrt{2}\\) sin (100t – \u03c0\/4)A and \u03c5 = 40 sin (100t) V. Calculate the resistance ?
        \nSolution:
        \ni = \\(\\sqrt{2}\\) sin (100t – \u03c0\/4)A
        \n(\u2235 i = i0<\/sub> sin(\u03c9t – \u03d5))
        \n\u03c5 = 40 sin (100t)V (\u2235 V = V0<\/sub>sin(\u03c9t))
        \ni0<\/sub> = \\(\\sqrt{2}\\), V0<\/sub> = 40, \u03c9 = 100, \u03d5 = \u03c0\/4
        \nR = \\(\\frac{\\mathrm{V}_0}{\\mathrm{i}_0} \\cos \\phi\\) = \\(\\frac{40}{\\sqrt{2}} \\cos \\frac{\\pi}{4}\\)
        \nR = \\(\\frac{40}{\\sqrt{2}} \\times \\frac{1}{\\sqrt{2}}\\)
        \nR = 20\u2126<\/p>\n

        Question 3.
        \nIn an AC circuit, a condenser, a resistor and a pure inductor are connected in series across an alternator (AC generator). If the voltages across them are 20 V, 35 V and 20 V respectively, find the voltage supplied by the alternator.
        \nSolution:
        \nVC<\/sub> = 20V, VR<\/sub> = 35V, VL<\/sub> = 20V
        \nV = \\(\\sqrt{\\mathrm{V}_{\\mathrm{R}}^2+\\left(\\mathrm{V}_{\\mathrm{L}}^2-\\mathrm{V}_{\\mathrm{C}}^2\\right)}\\)
        \nV = \\(\\sqrt{(35)^2+\\left(20^2-20^2\\right)}\\)
        \nV = \\(\\sqrt{35^2}\\)
        \nV = 35 Volt.<\/p>\n

        Question 4.
        \nAn AC circuit contains a resistance R, an inductance L and a capacitance C connected in series across an alternator of constant voltage and variable frequency. At resonant frequency, it is found that the inductive reactance, the capacitive reactance and the resistance are equal and the current in the circuit
        \nis i0<\/sub>. Find the current in the circuit at a frequency twice that of the resonant frequency.
        \nSolution:
        \nAt Resonance R = \u03c90<\/sub>L = \\(\\frac{1}{\\omega_0 C}\\)
        \n\"AP<\/p>\n

        \"AP<\/p>\n

        Question 5.
        \nA series resonant circuit contains L1<\/sub>, R1<\/sub> and C1<\/sub>. The resonant frequency is f. Another series resonant circuit contains L2<\/sub>, R2<\/sub> and C2<\/sub>. The resonant frequency is also f. If these two circuits are connected in series, calculate the resonant frequency.
        \nSolution:
        \nGiven that resonant frequency (f)
        \n= \\(\\frac{1}{2 \\pi \\sqrt{\\mathrm{L}_1 C_1}}\\) = \\(\\frac{1}{2 \\pi \\sqrt{\\mathrm{L}_2 \\mathrm{C}_2}}\\)
        \nL1<\/sub>C1<\/sub> = L2<\/sub>C2<\/sub>
        \nL1<\/sub> = \\(\\frac{\\mathrm{L}_2 \\mathrm{C}_2}{\\mathrm{C}_1}\\) —— (1)
        \nWhen these two circuits are connected in series
        \nThe total inductance L = L1<\/sub> + L2<\/sub>
        \nTotal capacitance is given by
        \n\\(\\frac{1}{\\mathrm{C}}\\) = \\(\\frac{1}{C_1}\\) + \\(\\frac{1}{C_2}\\) (or) C = \\(\\frac{\\mathrm{C}_1 \\mathrm{C}_2}{\\mathrm{C}_1+\\mathrm{C}_2}\\)
        \nThe resonant frequency of combined circuit is given by
        \n\"AP<\/p>\n

        Question 6.
        \nIn a series LCR circuit R = 200\u2126 and the voltage and the frequency of the mains supply is 200 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 45\u00b0. On taking out the inductor from the circuit the current leads the voltage by 45\u00b0. Calculate the power dissipated in the LCR circuit.
        \nSolution:
        \nR = 200\u2126, V = 200V, f = 50Hz
        \n\u03d5 = 45\u00b0, At resonance \u03c9L = \\(\\frac{1}{\\omega \\mathrm{C}}\\)
        \nPower dissipated (P) = Vi = \\(\\frac{\\mathrm{V}^2}{\\mathrm{R}}\\) (\u2235 i = \\(\\frac{V}{R}\\))
        \nP = \\(\\frac{(200)^2}{200}\\) = 200\u03c9<\/p>\n

        Question 7.
        \nThe primary of a transformer with primary to secondary turns ratio of 1: 2, is connected to an alternator of voltage 200 V. A current of 4A is flowing though the primary coil. Assuming that the transformer has no losses, find the secondary voltage and current are respectively.
        \nSolution:
        \nGiven that \\(\\frac{\\mathbf{N}_1}{\\mathbf{N}_2}\\) = \\(\\frac{1}{2}\\)
        \nE1<\/sub> = 200V, i1<\/sub> = 4A<\/p>\n

        i)
        \nTransformer ratio = \\(\\frac{E_2}{E_1}\\) = \\(\\frac{\\mathrm{N}_2}{\\mathrm{~N}_1}\\)
        \n\\(\\frac{E_2}{200}\\) = \\(\\frac{2}{1}\\)
        \nE2<\/sub> = 400V<\/p>\n

        ii)
        \n\\(\\frac{\\mathrm{N}_2}{\\mathrm{~N}_1}\\) = \\(\\frac{i_1}{i_2}\\)
        \n\\(\\frac{2}{1}\\) = \\(\\frac{4}{i_2}\\)
        \ni2<\/sub> = 2A<\/p>\n

        Textual Exercises<\/span><\/p>\n

        Question 1.
        \nA 100\u2126 resistor is connected to a 220 V. 50 Hz ac supply.
        \n(a) What is the rms value of current in the circuit ?
        \n(b) What is the net power consumed over a full cycle ?
        \nSolution:
        \nGiven resistance R = 100\u2126
        \nVrms<\/sub> = 220V
        \nFrequency f = 50Hz<\/p>\n

        a) Current in the circuit
        \nIrms<\/sub> = \\(\\frac{V_{r m s}}{R}\\) = \\(\\frac{220}{100}\\) = 2.2A<\/p>\n

        b) Net power consumed in full cycle
        \nP = Vrms<\/sub> \u00d7 Irms<\/sub>
        \n= 220 \u00d7 2.2
        \n= 474 W<\/p>\n

        Question 2.
        \n(a) The peak voltage of an ac supply is 300 V. What is the rms voltage ?
        \n(b) The rms value of current in an ac circuit is 10A. What is the peak current ?
        \nSolution:
        \na) Given, peak value of voltage V0<\/sub> = 300V
        \nThe rms value of current Irms<\/sub> = 10A
        \nThe rms value of voltage
        \nVrms<\/sub> = \\(\\frac{\\mathrm{V}_0}{\\sqrt{2}}\\) = \\(\\frac{300}{\\sqrt{2}}\\) = 212.1V<\/p>\n

        b) Using the formula
        \nIrms<\/sub> = \\(\\frac{\\mathrm{I}_0}{\\sqrt{2}}\\)
        \nThe peak value of current
        \nI0<\/sub> = \\(\\sqrt{2} \\times \\mathrm{I}_{\\mathrm{rms}}\\)
        \n= \\(\\sqrt{2}\\) \u00d7 10 = 1.414 \u00d7 10
        \nI0<\/sub> = 14.14 A<\/p>\n

        Question 3.
        \nA 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of current in the circuit.
        \nSolution:
        \nGiven inductance
        \nL = 44mH = 44 \u00d7 10-3<\/sup>H
        \nVrms<\/sub> = 220V
        \nFrequency of Inductor f = 50Hz
        \nInductive resistance XL<\/sub> = 2\u03c0fL
        \n= 2 \u00d7 3.14 \u00d7 50 \u00d7 44 \u00d7 10-3<\/sup>
        \n= 13.83\u2126
        \nThe rms value of current in the circuit
        \nIrms<\/sub> = \\(\\frac{V_{\\text {rms }}}{X_L}\\) = \\(\\frac{220}{13.83}\\)15.9A<\/p>\n

        Question 4.
        \nA 60 \u03bcF capacitor is connected to a 110V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
        \nSolution:
        \nGiven, capacitance of the capacitor C
        \n= 60\u03bcF
        \n= 60 \u00d7 10-6<\/sup>F
        \nVrms<\/sub> = 110
        \nFrequency of AC supply f = 60Hz
        \nCapacitive reactance
        \nXC<\/sub> = \\(\\frac{1}{2 \\pi \\mathrm{fC}}\\) = \\(\\frac{1}{2 \\times 3.14 \\times 60 \\times 60 \\times 10^{-6}}\\)
        \n= 44.23\u2126
        \nThe rms value of current in the circuit
        \nVrms<\/sub> = \\(\\frac{\\mathrm{V}_{\\mathrm{rms}}}{\\mathrm{X}_{\\mathrm{C}}}\\) = \\(\\frac{110}{44.23}\\) = 2.49A<\/p>\n

        \"AP<\/p>\n

        Question 5.
        \nIn Exercises 3 and 4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
        \nSolution:
        \nIn Exercise 3 Average power P = Vrms<\/sub>, Irms<\/sub>. cos \u03d5<\/p>\n

        As we know that the phase difference between current and voltage in case of inductor is 90\u00b0.
        \nP = Vrms<\/sub>. Irms<\/sub>. cos 90\u00b0 = 0
        \nIn Exercise 4 Average power
        \nP = Vrms<\/sub>. Irms<\/sub>. cos \u03d5
        \nWe know that the phase difference between current and voltage in case of capacitor is 90\u00b0
        \nP = Vrms<\/sub>. Irms<\/sub>. cos 90\u00b0 = 0<\/p>\n

        Question 6.
        \nObtain the resonant frequency \u03c9r<\/sub> of a series LCR circuit with L = 2.0H. C = 32 \u03bcF and R = 10\u2126. What is the Q-value of this circuit ?
        \nSolution:
        \nGiven, L = 2H, C = 32\u03bcF, R = 10\u2126
        \nResonant angular frequency
        \n\"AP<\/p>\n

        Question 7.
        \nA charged 30 \u03bcF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit ?
        \nSolution:
        \nCapacitance of capacitor C = 30\u03bcF = 30 \u00d7 10-6<\/sup>F
        \nInductance L = 27mH = 27 \u00d7 10-3<\/sup>H
        \nFor free oscillations, the angular frequency should be resonant frequency. Resonant angular frequency of oscillation of the circuit
        \n\"AP<\/p>\n

        Question 8.
        \nSuppose the initial charge on the capacitor in Exercise 7 is 6 mC. What is the total energy stored in the circuit initially ? What is the total energy at later time ?
        \nAnswer:
        \nGiven, charge on the capacitor Q = 6mC = 6 \u00d7 10-3<\/sup>
        \nC = 30\u03bcF = 30 \u00d7 10-6<\/sup>F
        \nEnergy stored in the circuit
        \nE = \\(\\frac{Q^2}{2 C}\\) = \\(\\frac{\\left(6 \\times 10^{-3}\\right)^2}{2 \\times 30 \\times 10^{-2}}\\) = \\(\\frac{36}{60}\\) = 0.6J
        \nAfter some time, the energy is shared between C and L, but the total energy remains constant.
        \nSo, we assume that there is no loss of energy.<\/p>\n

        Question 9.
        \nA series LCR circuit with R = 20\u2126, L = 1.5 H and C = 35\u00b5F is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle ?
        \nGiven resistance R = 200\u2126
        \nInductance L = 1.5H, capacitance C = 35\u00b5F = 35 \u00d7 10-6<\/sup>F and voltage Vrms<\/sub> = 200V
        \nWhen, the frequency of the supply equal to the natural frequency of the circuit, this is the condition of resonance. At the condition of resonance, Impedance Z = R = 20\u2126
        \nThe rms value of current in the circuit
        \nIrms<\/sub> = \\(\\frac{V_{r m s}}{Z}\\) = \\(\\frac{200}{20}\\) = 10A
        \n\u03d5 = 0\u00b0
        \nPower transferred to the circuit in one complete cycle.
        \nP = Irms<\/sub> Vrms<\/sub> cos \u03d5
        \n= 10 \u00d7 200 \u00d7 cos0\u00b0 = 2000 \u03c9 = 2K\u03c9<\/p>\n

        Question 10.
        \nA radio can tune over the frequency range o a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 \u03bcH, what must be the range of its variable capacitor?<\/p>\n

        (Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.)
        \nSolution:
        \nGiven, minimum frequency
        \nf1<\/sub> = 800KHz = 8 \u00d7 105<\/sup>HZ
        \nInductance
        \nL = 200\u03bcH = 200 \u00d7 10-5<\/sup>H = 2 \u00d7 10-4<\/sup>H.
        \nMaximum frequency
        \nf2<\/sub> = 1200 KHz = 12 \u00d7 105<\/sup>Hz for tuning, the natural frequency is equal to the frequency of oscillations that means it is the case of resonance.
        \nFrequency of oscillations f = \\(\\frac{1}{2 \\pi \\sqrt{L C}}\\)
        \nFor capacitance C1<\/sub>, f1<\/sub> = \\(\\frac{1}{2 \\pi \\sqrt{L C_i}}\\)
        \nC1<\/sub> = \\(\\frac{1}{4 \\pi^2 \\mathrm{f}_1^2 \\mathrm{~L}}\\)
        \n= \\(\\frac{1}{4 \\times 3.14 \\times 3.14 \\times\\left(8 \\times 10^5\\right)^2 \\times 2 \\times 10^{-4}}\\)
        \n= 197.7 \u00d7 10-12<\/sup>F
        \n= 197.7PF
        \nfor capacitance C2<\/sub>, f2<\/sub> = \\(\\frac{1}{2 \\pi \\sqrt{L C_2}}\\)
        \nC2<\/sub> = \\(\\frac{1}{4 \\pi^2 f_1^2 \\mathrm{C}}\\)
        \n= \\(\\frac{1}{4 \\times 3.14 \\times 3.14 \\times\\left(12 \\times 10^5\\right)^2 \\times 2 \\times 10^{-4}}\\)
        \n= 87.8 \u00d7 10-12<\/sup>F = 87.8pF
        \nThus, the range of capacitor is 87.8pF to 197.7pF.<\/p>\n

        Question 11.
        \nFigure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80\u03bcF, R = 40\u2126.
        \n\"AP
        \n(a) Determine the source frequency which drives the circuit in resonance.
        \n(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
        \n(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
        \nSolution:
        \nGiven, the rms value of voltage
        \nVrms<\/sub> = 230V
        \nInductance L = 5H
        \nCapacitance C = 80\u03bcf = 80 \u00d7 10-6<\/sup>F
        \nResistance R = 40\u2126<\/p>\n

        a) For resonance frequency of circuit
        \n\u03c9r<\/sub> = \\(\\frac{1}{\\sqrt{\\mathrm{LC}}}\\) = \\(\\frac{1}{\\sqrt{5 \\times 80 \\times 10^{-6}}}\\) = 50 rad\/s
        \nSource frequency at resonance, then
        \n\u03c50<\/sub> = \\(\\frac{\\omega_0}{2 \\pi}\\) = \\(\\frac{50}{2 \\times 3.14}\\) = 7.76 Hz<\/p>\n

        b) At the resonant frequency, XL<\/sub> = XC<\/sub>
        \nSo, impedance of the circuit Z = R
        \n\u2234 Impedance Z = 40\u2126
        \nThe rms value of current in the circuit
        \nIrms<\/sub> = \\(\\frac{V_{r m s}}{Z}\\) = \\(\\frac{230}{40}\\) = 5.75A
        \nAmplitude of current I0<\/sub> = Irms<\/sub>\\(\\sqrt{2}\\)
        \n= 5.75 \u00d7 \\(\\sqrt{2}\\)
        \n= 8.13A<\/p>\n

        c) The rms potential drop across L
        \nVL<\/sub> = Irms<\/sub> XL<\/sub>
        \n= Irms<\/sub> \u03c9r<\/sub>L = 5.75 \u00d7 50 \u00d7 5 = 1437.5V
        \nThe rms potential drop across R
        \nVR<\/sub> = Irms<\/sub> = R = 5.75 \u00d7 40 = 230V
        \nThe rms potential drop across C,
        \n\"AP
        \nPotential drop across LC combinations
        \n= Irms<\/sub>(XL<\/sub> – XC<\/sub>)
        \n= Irms<\/sub>(XL<\/sub> – XL<\/sub>) = 0 (\u2235 XL<\/sub> – XC<\/sub>)<\/p>\n

        Additional Exercises<\/span><\/p>\n

        Question 1.
        \nAn LC circuit contains a 20 mH inductor and a 50\u03bcF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0.
        \n(a) What is the total energy stored initially ? Is it conserved during LC oscillations ?
        \n(b) What is the natural frequency of the circuit ?
        \n(c) At What time is the energy stored
        \n(i) Completely electrical (i.e., stored in the capacitor) ?
        \n(ii) completely magnetic (i.e., stored in the inductor) ?
        \n(d) At what times is the total energy shared equally between the inductor and the capacitor ?
        \n(e) As a resistor is inserted in the circuit, how much energy is eventually dissipated as heat ?
        \nSolution:
        \nGiven Inductance L = 20mH = 20 \u00d7 10-3<\/sup>H
        \nCapacitance of capacitor
        \nC = 50\u03bcf = 50 \u00d7 10-6<\/sup>F
        \nInitial charge on the capacitor,
        \nQi<\/sub> = 10mc = 10 \u00d7 103<\/sup>C<\/p>\n

        a) The total energy stored across the capacitor initially,
        \n\"AP
        \nYes, this energy is conserved during LC oscillations.<\/p>\n

        b) To get the natural frequency or resonant frequency,
        \nfr<\/sub> = \\(\\frac{1}{2 \\pi \\sqrt{L C}}\\)
        \n= \\(\\frac{1}{2 \\times 3.14 \\times \\sqrt{20 \\times 10^{-3} \\times 50 \\times 10^{-6}}}\\)
        \n= \\(\\frac{7 \\times 10^3}{44}\\) = 159.2 Hz.
        \nThe natural frequency of the circuit
        \n\u03c9 = 2\u03c0V = 2\u03c0 \u00d7 159.2
        \n= 999.78 \u2248 1000 = 103<\/sup> rad\/s<\/p>\n

        c)<\/p>\n

        i) Let at any instant the energy stored is completely electrical the charge on the capacitor Q = Q0<\/sub> cos \u03c9t
        \nQ = Q0<\/sub> cos \\(\\frac{2 \\pi}{T} \\cdot t\\) —– (1)
        \nQ is maximum as it is equal to Q0<\/sub>, only if cos \\(\\frac{2 \\pi}{T}\\) t = \u00b11 = cosn\u03c0t or t
        \n= \\(\\frac{\\mathrm{nT}}{2}\\), where n = 1, 2, 3. ..
        \nt = 0, T\/2, T, 3T\/2, ………..
        \nThus, the energy stored is completely electrical at t = 0, T\/2, T, 3\/2……….<\/p>\n

        ii) Let at any instant, the energy stored is completely magnetic as when the electrical energy across the capacitor is zero.
        \nq = 0
        \nQ = Q0<\/sub>cos \\(\\frac{2 \\pi t}{T}\\) = 0 (from eq(i))
        \n\u2234 cos \\(\\frac{2 \\pi}{T} \\cdot t\\) = 0 = \\(\\frac{\\mathrm{n} \\pi}{2}\\) or t = \\(\\frac{\\mathrm{nT}}{4}\\) = cos.
        \nIt happens if t = T\/4, 3T\/4, \\(\\frac{\\mathrm{5T}}{4}\\),………..
        \nThus, the energy stored is completely magnetic at
        \nt = T\/4, 3T\/4, 5T\/4………..<\/p>\n

        d) Equal sharing of energy between inductor and capacitor means the energy stored in capacitor = \\(\\frac{1}{2}\\) \u00d7 Maximum energy
        \n\"AP
        \nor cos(2n + 1)\u03c0\/4 = cos \\(\\frac{2 \\pi \\mathrm{t}}{\\mathrm{T}}\\)
        \n\\(\\frac{(2 n+1) \\pi}{4}\\) = \\(\\frac{2 \\pi t}{T}\\)
        \nt = T\/8(2n + 1) (n = 0, 1, 2, 3, ……..)
        \nHence the energy will be shared half on capacitor and half on inductor.
        \nt = \\(\\frac{T}{8}\\), \\(\\frac{3 \\mathrm{~T}}{8}\\), \\(\\frac{5 \\mathrm{~T}}{8}\\), ………..<\/p>\n

        e) As a resistor is inserted in the circuit, all of the energy loss during heating. Energy loss = 1J. The oscillations becomes damped and becomes disappear after sometime as the total energy loss in the form of heat.<\/p>\n

        \"AP<\/p>\n

        Question 2.
        \nA coil of inductance 0.50 H and resistance 100\u2126 is connected to a 240 V, 50Hz ac supply.
        \na) What is the maximum current in the coil ?
        \nb) What is the time lag between the voltage maximum and the current maximum ?
        \nSolution:
        \nGiven, Inductance L = 0.50H
        \nResistance R = 100\u2126
        \nThe rms value of voltage
        \nVrms<\/sub> = 240V, f = 50Hz<\/p>\n

        a) Impedance of circuit
        \n\"AP
        \nthe rms value of current
        \nIrms<\/sub> = \\(\\frac{\\mathrm{V}_{\\mathrm{rms}}}{\\mathbf{Z}}\\) = \\(frac{240}{31400.15}\\) = 0.00764
        \nthe maximum value of current in the circuit
        \nI0<\/sub> = \\(\\sqrt{2}\\) Irms<\/sub> = 1.414 \u00d7 0.00764 = 1.824A<\/p>\n

        b) Using the formula of time lag,
        \nt = \\(\\frac{\\phi}{\\omega}\\)
        \ntan \u03d5 = \\(\\frac{X_L}{R}\\) = \\(\\frac{\\omega \\mathrm{L}}{\\mathrm{R}}\\) = \\(\\frac{2 \\pi \\mathrm{fL}}{\\mathrm{R}}\\) = \\(\\frac{2 \\times 3.14 \\times 50 \\times 0.50}{100}\\)
        \n\u03d5 = tan-6<\/sup>(1.571) = 57.5
        \n\"AP
        \nTime lag t = \\(\\frac{\\phi}{\\omega}\\) = \\(\\frac{57.5 \\pi}{180 \\times 2 \\pi \\mathrm{f}}\\)
        \n= \\(\\frac{57.5}{180 \\times 2 \\times 50}\\)
        \n= 3.19 \u00d7 10-3<\/sup>S
        \nThus, the time lag between the voltage maximum and the current maximum is 3.19 \u00d7 10-3<\/sup>S<\/p>\n

        Question 3.
        \nObtain the answers (a) to (b) in Exercise 13 if the circuit is connected to a high frequency supply (240V, 10 kHz). Hence, explain the statement that at very high frequency an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
        \nAnswer:
        \nGiven frequency f = 10kHz = 104<\/sup>Hz
        \nthe rms value of voltage Vrms<\/sub> = 240v
        \nfrom Exercise 13
        \nResistance R = 100\u2126
        \nInductance L = 0.5H
        \n\"AP<\/p>\n

        Question 4.
        \nA 100\u03bcF capacitor in series with a 40\u2126 resistance is connected to a 110V. 60 Hz supply.
        \n(a) What is the maximum current in the circuit?
        \n(b) What is the time lag between the current maximum and the voltage maximum?
        \nAnswer:
        \nGiven capacitance of capacitor
        \nC = 100\u00b5F= 100 \u00d7 10-6<\/sup>F
        \nResistance R = 40\u2126
        \nthe rms value of voltage Vrms<\/sub> = 110V
        \nFrequency f = 60Hz<\/p>\n

        (a) Impedance Z
        \n\"AP<\/p>\n

        (b) Time lag (t) = \\(\\frac{\\phi}{\\omega}\\)
        \n\"AP
        \nThus, the time lag between the voltage maximum and the current maximum is 1.55 \u00d7 10-3<\/sup>S.<\/p>\n

        Question 5.
        \nObtain the answers to (a) and (b) in Exercise 15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
        \nAnswer:
        \nGiven, the rms value of voltage,
        \nVrms<\/sub> = 110V
        \nThe frequency of capacitor
        \nf = 12kHz = 12000 Hz.
        \nCapacitance of conductor C = 10-4<\/sup>F.
        \nResistance R = 40\u2126
        \nCapacitive Resistance
        \nXC<\/sub> = \\(\\frac{1}{2 \\pi \\mathrm{fC}}\\) = \\(\\frac{1}{2 \\times 3.14 \\times 12000 \\times 10^{-4}}\\)
        \n= 0.133\u2126
        \nThe rms value of current
        \n\"AP
        \nThe maximum value of current,
        \nI0<\/sub> = \\(\\sqrt{2}\\) Irms<\/sub>
        \n= 1.414 \u00d7 2.75
        \n= 3.9A
        \nHere, the value of XC<\/sub> is very small, so term containing C is negligible.
        \ntan \u03d5 = \\(\\frac{1}{\\omega \\mathrm{CR}}\\)
        \n= \\(\\frac{1}{2 \\times 3.14 \\times 12000 \\times 10^{-4} \\times 40}\\)
        \n= \\(\\frac{1}{96 \\pi}\\)
        \nIt is very very small.
        \nIn DC circuits, \u03c9 = 0
        \nXC<\/sub> = \\(\\frac{1}{\\omega C}\\) = \u221e
        \nSo, it behaves like an open circuit.<\/p>\n

        Question 6.
        \nKeeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 11 for this frequency.
        \nSolution:
        \nAs they are corrected in the parallel combination
        \n\\(\\frac{1}{Z}\\) = \\(\\frac{1}{R}\\) + (\\(\\frac{1}{X_L}\\) + \\(\\frac{1}{X_C}\\))
        \nAs the reactance (XC<\/sub> – XL<\/sub>) is perpendicular to the ohmic resistance R, therefore we can write as
        \n\"AP
        \nThat means \\(\\frac{1}{Z}\\) = minimum and thus Z = maximum. As Z is maximum, current will be minimum, current through inductor
        \n\"AP<\/p>\n

        Question 7.
        \nA circuit containing a 80 mH inductor and a 60\u03bcF capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible.
        \n(a) Obtain the current amplitude and rms values.
        \n(b) Obtain the rms values of potential drops across each element.
        \n(c) What is the average power transferred to the inductor?
        \n(d) What is the average power transferred to the capacitor?
        \n(e) What is the total average power absorbed by the circuit? [\u2018Average\u2019 implies \u2018averaged over one cycle\u2019]
        \nAnswer:
        \nGiven Inductance
        \nL = 80mH = 80 \u00d7 10-3<\/sup>H
        \nCapacitance of capacitor
        \nC = 60\u03bcF = 60 \u00d7 10-6<\/sup>F .
        \nThe rms value of voltage
        \nVrms<\/sub> = 230V
        \nFrequency f = 50Hz; Resistance R = 0<\/p>\n

        a) Impedance of circuit
        \n\"AP
        \nAs ZCO, means XL<\/sub> < XC<\/sub>, emf lags by current by 90\u00b0
        \nThe rms value of current
        \nIrms<\/sub> = \\(\\frac{\\mathrm{V}_{\\mathrm{rms}}}{\\mathrm{Z}}\\) = \\(\\frac{230}{27.95}\\) = 8.29A
        \nThe maximum value of current
        \nI0<\/sub> = \\(\\sqrt{2} I_{\\text {rms }}\\) = 1.414 \u00d7 8.29 = 11.64A<\/p>\n

        b) Potenti\u00e1l drop acr\u00f5ss L
        \nVrms<\/sub>L = Irms<\/sub> \u00d7 XL<\/sub> = 8.29 \u00d7 2 \u00d7 3.14 \u00d7 50 \u00d7 80 \u00d7 10-3<\/sup>
        \n= 208.25 V
        \nPotential drop across C
        \nVrms<\/sub>C = Irms<\/sub> \u00d7 XC<\/sub>
        \n= 8.29 \u00d7 \\(\\frac{1}{2 \\times 3.14 \\times 50 \\times 60 \\times 10^{-6}}\\)
        \n= 440.02 V
        \n\u2234 Applied rms voltage = VC<\/sub> – VL<\/sub>
        \n= 440 – 208.25 = 231.75V<\/p>\n

        c) Average power transferred to the inductor
        \nP = Irms<\/sub> Vrms<\/sub>. cos\u03d5
        \nAs the phase difference is 90\u00b0, So P = 0<\/p>\n

        d) Average power transferred to the capacitor.
        \nP = Irms<\/sub> Vrms<\/sub>. cos\u03d5
        \nAs the phase difference is 90\u00b0, So P = 0<\/p>\n

        e) As there is no resistance in the circuit, so average power is equal to sum of average power due to inductor and capacitor. That means the average power consumed is zero.<\/p>\n

        \"AP<\/p>\n

        Question 8.
        \nSuppose the circuit in Exercise 18 has a resistance of 15\u2126. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
        \nSolution:
        \nGiven, the rms value of voltage V = 230V
        \nVrms<\/sub> = 230V
        \nResistance R = 15\u2126
        \nFrequency f = 50Hz
        \nAverage power across inductor and capacitor is zero as the phase difference between current and voltage is 90\u00b0.
        \nTotal power absorbed = power absorbed in resistor, Pav<\/sub>
        \n\"AP
        \nTotal power absorbed = 790.6W<\/p>\n

        Question 9.
        \nA series LCR circuit with L = 0.12 H, C = 480 nF, R = 23\u2126 is connected to a 230 V variable frequency supply.
        \na) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
        \nb) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
        \nc) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency ? what is the current amplitude at these frequencies ?
        \nd) What is the Q-factor of the given circuit ?
        \nAnswer:
        \nInductance L = 0.12H,
        \nCapacitance C = 480nF = 480 \u00d7 10-9<\/sup>F
        \nResistance R = 23\u2126
        \nThe rms value of voltage
        \nIrms<\/sub> = \\(\\frac{V_{r m s}}{Z}\\) = \\(\\frac{230}{23}\\) = 10A
        \nThe maximum value of current I0<\/sub> = \\(\\sqrt{2}\\)
        \nIrms<\/sub> = 1.414 \u00d7 10 = 14.14A.
        \nAt natural frequency, the current amplitude is maximum.
        \n\u03c9 = \\(\\frac{1}{\\sqrt{\\mathrm{LC}}} \\frac{1}{2 \\pi \\sqrt{0.12 \\times 480 \\times 10^{-9}}}\\) = 4166.6
        \n= 4167 rad\/s<\/p>\n

        b) Average power is maximum for resonance.
        \nPav(max)<\/sub> = I2<\/sup>rms<\/sub> R = 10 \u00d7 10 \u00d7 23 = 2300W.<\/p>\n

        c) Power transferred to the circuit is half the power at resonant frequency
        \n\u0394\u03c9 = \\(\\frac{\\mathrm{R}}{2 \\mathrm{~L}}\\) = \\(\\frac{23}{2 \\times 0.12}\\) = 95.83 rad\/s
        \n\u0394V = \\(\\frac{\\Delta \\mathrm{W}}{2 \\pi}\\) = 15.2Hz
        \nThe frequencies at which power transferred is half .
        \nV = V0<\/sub> \u00b1 \u0394V = 663.48 \u00b1 15.26
        \nSo, frequencies are 448.3Hz and 678.2Hz, the maximum current
        \nI = \\(\\frac{\\mathrm{I}_0}{\\sqrt{2}}\\) = \\(\\frac{14.14}{\\sqrt{2}}\\) = 10A<\/p>\n

        d) Q-factor = \\(\\frac{\\omega_{\\mathrm{r}}}{\\mathrm{R}}\\) = \\(\\frac{4166.7 \\times 0.12}{23}\\) = 21.74<\/p>\n

        Question 10.
        \nObtain the resonant frequency and Q-factor of series LCR circuit with L = 3.0 H, C = 27\u03bcF, and R = 10.4\u2126. It is desired to improved the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
        \nSolution:
        \nGiven, inductance L = 3H
        \nCapacitance of capacitor
        \nC = 27\u03bcF = 27 \u00d7 10-6<\/sup>F
        \nResistance R = 7.4\u2126
        \nthe resonant frequency of circuit
        \n\u03c9r<\/sub> = \\(\\frac{1}{\\sqrt{\\mathrm{LC}}}\\) = \\(\\frac{1}{\\sqrt{3 \\times 27 \\times 10^{-6}}}\\) = \\(\\frac{1000}{9}\\)
        \n= 111.1 rad\/s.
        \nQ-factor of a series LCR circuit
        \nQ-factor = \\(\\frac{\\omega_{\\mathrm{r}} \\mathrm{L}}{\\mathrm{R}}\\) = \\(\\frac{111.1 \\times 3}{7.4}\\) = 45.04
        \nTo reduce the full width at half by factor Q, we have to reduce the value of R as R\/2
        \n\\(\\frac{R}{2}\\) = \\(\\frac{7.4}{2}\\) = 3.7\u2126<\/p>\n

        Question 11.
        \nAnswer the following questions :
        \na) In any ac circuit, is the applied Instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit ? Is the same true for rms voltage ?
        \nAnswer:
        \nYes, the applied voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit
        \nNo, it is not true for rms voltage because there is some phase differences across different elements of circuits.<\/p>\n

        b) A capacitor is used in the primary circuit of an induction coil.
        \nAnswer:
        \nA capacitor is used in the primary circuit of an induction coil because when the circuit is broken, a large induced voltage is used up in charging the capacitor. So, the parking or any damages are avoided.<\/p>\n

        c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.
        \nSolution:
        \nAs we know that
        \nXC<\/sub> = \\(\\frac{1}{2 \\pi f c}\\), XL<\/sub> = 2\u03c0fL
        \nf = 0, XC<\/sub> = \u221e, XL<\/sub> = 0<\/p>\n

        d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an Iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
        \nAnswer:
        \nWhen the choke coil is connected to dc, there is no change in the brightness. Because f = 0, XL<\/sub> = 6. So, no change in the brightness. In Ac, the choke offers impedance, so, it glows dim. As we insert an iron core the magnetic field increases and hence inductance increases.
        \nBA = LI = \u03d5
        \nL \u221d B
        \nSo, XL<\/sub> also increases and the brightness of bulb decreases.<\/p>\n

        e) Why is choke coil needed in the use of fluorescent tubes with ac mains ? Why can we not use an ordinary resistor instead of the choke coil ?
        \nAnswer:
        \nWe use the choke coil instead of resistance because the power loss across resistor is maximum while the power loss across choke is zero.
        \nFor resistor, \u03d5 = 0
        \nP = Irms<\/sub>. Vrms<\/sub>. cos \u03d5
        \n= Irms<\/sub>. Vrms<\/sub> = maximum
        \nFor Inductor \u03d5 = 90\u00b0
        \nP = Irms<\/sub>. Vrms<\/sub>. cos 90\u00b0 = 0<\/p>\n

        Question 12.
        \nA power transmission line feeds input power at 2300 V to a stepdown transformer with its primary windings having 4000 turns, what should be the number of turns in the secondary in order to get output power at 230 V ?
        \nSolution:
        \nGiven Primary voltage VP<\/sub> = 2300V NP<\/sub> = 4000 turns
        \nSecondary voltage VS<\/sub> = 230V
        \nUsing formula,
        \n\\(\\frac{\\mathrm{V}_{\\mathrm{S}}}{\\mathrm{V}_{\\mathrm{P}}}\\) = \\(\\frac{N_S}{N_P}\\)
        \n\\(\\frac{230}{2300}\\) = \\(\\frac{\\mathrm{N}_{\\mathrm{S}}}{4000}\\)
        \nNS<\/sub> = 400
        \n\u2234 thus, the number of turns in secondary are 400.<\/p>\n

        \"AP<\/p>\n

        Question 13.
        \nAt a hydroelectric power plant, the water, pressure head is at a height of 300 m and the water flow available is 100 m3<\/sup>s-1<\/sup>. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (q = 9.8 ms-2<\/sup>).
        \nAnswer:
        \nGiven, height of water h = 300m
        \nRate of flow of water V = 100m3<\/sup>\/s
        \nefficiency \u03b7 = 60%
        \ng = 9.8m\/s2<\/sup>.
        \nAs we know that input power is required to raised the water up to height
        \nh = 300 m
        \n\"AP
        \nSuppose, the power output is Pout<\/sub> which is equal to the power available from the plant. The efficiency of generator
        \n\u03b7 = \\(\\frac{P_{\\text {out }}}{P_{\\text {in }}}\\)
        \n\"AP
        \npout<\/sub> = 176.4MW = 1764 \u00d7 105<\/sup>W<\/p>\n

        Question 14.
        \nA small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power in 0.5 \u2126 per km. The town gets power from the line through a 4000 – 220V step-down transformer at a substation in the town.
        \n(a) Estimate the line power loss in the form of heat.
        \n(b) How much power must the plant supply, assuming there is negligible power loss due to leakage ?
        \n(c) Characterise the step up transformer at the plant.
        \nAnswer:
        \nGenerating power of Electric plant = 800 kW at V = 220V
        \nDistance = 15km,
        \nGenerating voltage = 440V
        \nResistance\/length = 0.50\u2126\/km
        \nPrimary voltage VP<\/sub> = 4000V
        \nSecondary voltage VS<\/sub> = 220V
        \n(a) Power = Ip<\/sub>. Vp<\/sub>
        \n800 \u00d7 1000 = Ip<\/sub> \u00d7 4000
        \nIp<\/sub> = 200A
        \nLine power loss in form of heat
        \n= (Ip<\/sub>)2<\/sup> \u00d7 Resistance of line
        \n= (Ip<\/sub>)2<\/sup> \u00d7 0.5 \u00d7 15 \u00d7 2
        \n= (200)2<\/sup> \u00d7 0.5 \u00d7 15 \u00d7 2
        \n= 60 \u00d7 104<\/sup>W
        \n= 600KW<\/p>\n

        (b) If there is no power loss due to leakage the plant supply should be =
        \n800 + 600 = 1400KW<\/p>\n

        (c) Voltage drop across the line
        \n= IP<\/sub>. R
        \n= 200 \u00d7 0.5 \u00d7 15 \u00d7 2
        \n= 3000V
        \nVoltage from transmission
        \n= 3000 + 4000 = 7000V
        \nAs it is given that the power generated at 440V
        \nSo, the step-up transformer needed at the plant is 440V – 7000V<\/p>\n

        Question 15.
        \nDo the same exercise as above with the replacement of the earlier transformer by a 40,000-220V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred ?
        \nAnswer:
        \nGiven primary voltage VP<\/sub> = 40,000V
        \nLet the current in primary is IP<\/sub>
        \n\u2234 VP<\/sub>IP<\/sub> = P
        \n800 \u00d7 1000 = 40000 \u00d7 IP<\/sub>
        \nIP<\/sub> = 20A
        \na) Line power loss = (IP<\/sub>)2<\/sup> \u00d7 R
        \n= (20)2<\/sup> \u00d7 2 \u00d7 0.5 \u00d7 15
        \n= 6000W = 6KW<\/p>\n

        b) Power supply by plant
        \n= 800 + 6 = 806KW<\/p>\n

        Voltage drop of line = IP<\/sub>.R
        \n= 20 \u00d7 2 \u00d7 0.5 \u00d7 15
        \n= 300V
        \nVoltage for transmission
        \n= 40000 + 300 = 40300V
        \nPower loss at higher voltage
        \n= \\(\\frac{6}{800}\\) \u00d7 100 = 0.74%
        \nPower loss at lower voltage
        \n= \\(\\frac{600}{1400}\\) \u00d7 100 = 42.8%
        \nHence, the power loss is minimum at higher voltage. So, the high voltage transmission is preferred.<\/p>\n","protected":false},"excerpt":{"rendered":"

        Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 10th Lesson Alternating Current Textbook Questions and Answers. AP Inter 2nd Year Physics Study Material 10th Lesson Alternating Current Very Short Answer Questions Question 1. A transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the secondary if the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35772"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35772"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35772\/revisions"}],"predecessor-version":[{"id":35773,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35772\/revisions\/35773"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35772"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35772"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35772"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}