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NEET-XII-Physics

07: Alternating Current

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    Chapter 7 - Alternating Current
  • Qstn #1
    A 100 ``\Omega`` resistor is connected to a 220 V, 50 Hz ac supply.
    Ans : Resistance of the resistor, R = 100 ``\Omega``

    Supply voltage, V = 220 V

    Frequency, ν = 50 Hz
  • #1-a
    What is the rms value of current in the circuit?
    Ans : The rms value of current in the circuit is given as:


  • #1-b
    What is the net power consumed over a full cycle?
    Ans : The net power consumed over a full cycle is given as:

    P = VI

    = 220 × 2.2 = 484 W
  • #2
  • #2-a
    The peak voltage of an ac supply is 300 V. What is the rms voltage?
    Ans : Peak voltage of the ac supply, V0 = 300 V

    Rms voltage is given as:


  • #2-b
    The rms value of current in an ac circuit is 10 A. What is the peak current?
    Ans : Therms value of current is given as:

    I = 10 A

    Now, peak current is given as:


  • Qstn #3
    A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
    Ans : Inductance of inductor, L = 44 mH = 44 × 10-3 H

    Supply voltage, V = 220 V

    Frequency, ν = 50 Hz

    Angular frequency, ω=

    Inductive reactance, XL = ω L

    Rms value of current is given as:



    Hence, the rms value of current in the circuit is 15.92 A.
  • Qstn #4
    A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
    Ans : Capacitance of capacitor, C = 60 μF = 60 × 10-6 F

    Supply voltage, V = 110 V

    Frequency, ν = 60 Hz

    Angular frequency, ω=

    Capacitive reactance



    Rms value of current is given as:



    Hence, the rms value of current is 2.49 A.
  • Qstn #5
    In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
    Ans : In the inductive circuit,

    Rms value of current, I = 15.92 A

    Rms value of voltage, V = 220 V

    Hence, the net power absorbed can be obtained by the relation,

    P = VI cos Φ

    Where,

    Φ = Phase difference between V and I

    For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.

    Hence, P = 0 i.e., the net power is zero.

    In the capacitive circuit,

    Rms value of current, I = 2.49 A

    Rms value of voltage, V = 110 V

    Hence, the net power absorbed can ve obtained as:

    P = VI Cos Φ

    For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.

    Hence, P = 0 i.e., the net power is zero.
  • Qstn #6
    Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 ``\Omega``. What is the Q-value of this circuit?
    Ans : Inductance, L = 2.0 H

    Capacitance, C = 32 μF = 32 × 10-6 F

    Resistance, R = 10 ``\Omega``

    Resonant frequency is given by the relation,



    Now, Q-value of the circuit is given as:



    Hence, the Q-Value of this circuit is 25.
  • Qstn #7
    A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
    Ans : Capacitance, C = 30μF = 30×10-6F

    Inductance, L = 27 mH = 27 × 10-3 H

    Angular frequency is given as:



    Hence, the angular frequency of free oscillations of the circuit is 1.11 × 103 rad/s.
  • Qstn #8
    Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
    Ans : Capacitance of the capacitor, C = 30 μF = 30×10-6 F

    Inductance of the inductor, L = 27 mH = 27 × 10-3 H

    Charge on the capacitor, Q = 6 mC = 6 × 10-3 C

    Total energy stored in the capacitor can be calculated by the relation,



    Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.
  • Qstn #9
    A series LCR circuit with R = 20 ``\Omega``, L = 1.5 H and C = 35 μF 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?
    Ans : At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.

    Resistance, R = 20 ``\Omega``

    Inductance, L = 1.5 H

    Capacitance, C = 35 μF = 30 × 10-6 F

    AC supply voltage to the LCR circuit, V = 200 V

    Impedance of the circuit is given by the relation,



    At resonance,



    Current in the circuit can be calculated as:



    Hence, the average power transferred to the circuit in one complete cycle= VI

    = 200 × 10 = 2000 W.
  • Qstn #10
    A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?

    [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.]
    Ans : The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.

    Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz

    Upper tuning frequency, ν2 = 1200 kHz = 1200× 103 Hz

    Effective inductance of circuit L = 200 μH = 200 × 10-6 H

    Capacitance of variable capacitor for ν1 is given as:

    C1

    Where,

    ω1 = Angular frequency for capacitor C1





    Capacitance of variable capacitor for ν2,

    C2

    Where,

    ω2 = Angular frequency for capacitor C2





    Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.
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