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NEET-XII-Physics

39: Alternating Current

with Solutions - page 4
Qstn# iv-14-a Prvs-QstnNext-Qstn
  • #14-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 : 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-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: