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NEET-XII-Physics

39: Alternating Current

with Solutions - page 3
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  • #14
    In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, εrms = 50 V and ν = 50/π Hz. Find (a) the rms current in the circuit and (b) the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.
    Ans : Given:
    Resistance in series LCR circuit, R = 300 Ω
    Capacitance in series LCR circuit, C = 20 μF= 20 × 10-6 F
    Inductance in series LCR circuit, L = 1 Henry
    RMS value of voltage, εrms = 50 V
    Frequency of source, f = 50/`` \,\mathrm{\,\pi \,}`` Hz
    Reactance of the inductor (XL) is given by,
    XL= `` \omega L`` = 2`` \,\mathrm{\,\pi \,}``fL
    `` \Rightarrow ``XL = 2`` \times \,\mathrm{\,\pi \,}\times \frac{50}{\,\mathrm{\,\pi \,}}\times 1`` = 100 `` \,\mathrm{\,\Omega \,}``
    Reactance of the capacitance `` \left({X}_{C}\right)`` is given by,
    `` {X}_{C}=\frac{1}{\omega C}`` = `` \frac{1}{2\,\mathrm{\,\pi \,}fC}``
    `` \Rightarrow ``XC = `` \frac{1}{2\,\mathrm{\,\pi \,}\times {\displaystyle \frac{50}{\,\mathrm{\,\pi \,}}}\times 20\times {10}^{-6}}``
    `` \Rightarrow `` XC = `` 500`` `` \,\mathrm{\,\Omega \,}`` (a) Impedance of an LCR circuit `` \left(Z\right)`` is given by,
    `` Z=\sqrt{{R}^{2}+{\left({X}_{\,\mathrm{\,C\,}}-{X}_{\,\mathrm{\,L\,}}\right)}^{2}}``
    `` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(500-100\right)}^{2}}``
    `` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}``
    `` \Rightarrow Z=500``
    `` ``
    RMS value of current `` \left({I}_{rms}\right)`` is given by,
    `` {I}_{rms}=\frac{{\epsilon }_{rms}}{Z}``
    `` \Rightarrow ``Irms = `` \frac{50}{500}``
    `` \Rightarrow ``Irms = `` 0.1\,\mathrm{\,A\,}`` (b) Potential across the capacitor `` \left({V}_{C}\right)`` is given by,
    VC = Irms × XC
    `` \Rightarrow ``VC = 0.1 × 500 = 50 V
    Potential difference across the resistor `` \left({V}_{R}\right)`` is given by,
    VR = Irms × R
    `` \Rightarrow ``VR = 0.1 × 300 = 30 V
    Potential difference across the inductor `` \left({V}_{L}\right)`` is given by,
    VL = Irms × XL
    `` \Rightarrow `` VL= 0.1 × 100 = 10 V
    R.M.S potential = 50 V
    Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
    Sum of the potential drops > RMS potential applied
    Page No 330:
  • #14-a
    the rms current in the circuit and
    Ans : Impedance of an LCR circuit `` \left(Z\right)`` is given by,
    `` Z=\sqrt{{R}^{2}+{\left({X}_{\,\mathrm{\,C\,}}-{X}_{\,\mathrm{\,L\,}}\right)}^{2}}``
    `` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(500-100\right)}^{2}}``
    `` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}``
    `` \Rightarrow Z=500``
    `` ``
    RMS value of current `` \left({I}_{rms}\right)`` is given by,
    `` {I}_{rms}=\frac{{\epsilon }_{rms}}{Z}``
    `` \Rightarrow ``Irms = `` \frac{50}{500}``
    `` \Rightarrow ``Irms = `` 0.1\,\mathrm{\,A\,}``
  • #14-b
    the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.
    Ans : Potential across the capacitor `` \left({V}_{C}\right)`` is given by,
    VC = Irms × XC
    `` \Rightarrow ``VC = 0.1 × 500 = 50 V
    Potential difference across the resistor `` \left({V}_{R}\right)`` is given by,
    VR = Irms × R
    `` \Rightarrow ``VR = 0.1 × 300 = 30 V
    Potential difference across the inductor `` \left({V}_{L}\right)`` is given by,
    VL = Irms × XL
    `` \Rightarrow `` VL= 0.1 × 100 = 10 V
    R.M.S potential = 50 V
    Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
    Sum of the potential drops > RMS potential applied
    Page No 330: