HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

39: Alternating Current

with Solutions - page 3
Qstn# iv-16 Prvs-QstnNext-Qstn
  • #16
    An inductance of 2.0 H, a capacitance of 18μF and a resistance of 10 kΩ are connected to an AC source of 20 V with adjustable frequency. (a) What frequency should be chosen to maximise the current in the circuit? (b) What is the value of this maximum current?
    Ans : Given:
    Inductance of inductor, L = 2.0 H
    Capacitance of capacitor, C = 18 μF
    Resistance of resistor, R = 10 kΩ
    Voltage of AC source, E = 20 V (a) In an LCR circuit, current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
    XL = XC
    `` \Rightarrow \omega L=\frac{1}{\omega C}``
    `` \Rightarrow {\omega }^{2}=\frac{1}{LC}=\frac{1}{2\times 18\times {10}^{-6}}``
    `` \Rightarrow {\omega }^{2}=\frac{{10}^{6}}{36}``
    `` \Rightarrow \omega =\frac{{10}^{3}}{6}``
    `` \Rightarrow 2\,\mathrm{\,\pi \,}f=\frac{{10}^{3}}{6}``
    `` \Rightarrow f=\frac{1000}{6\times 2\,\mathrm{\,\pi \,}}=26.539\,\mathrm{\,Hz\,}``
    `` =27\,\mathrm{\,Hz\,}`` (b) At resonance, reactance is minimum.
    Minimum Reactance, Z = R
    Maximum current (I) is given by,
    `` I=\frac{E}{R}``
    `` \Rightarrow I\mathit{=}\frac{20}{10\times {10}^{3}}``
    `` \Rightarrow I=\frac{2\,\mathrm{\,A\,}}{{10}^{3}}=2\,\mathrm{\,mA\,}``
    Page No 330:
  • #16-a
    What frequency should be chosen to maximise the current in the circuit?
    Ans : In an LCR circuit, current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
    XL = XC
    `` \Rightarrow \omega L=\frac{1}{\omega C}``
    `` \Rightarrow {\omega }^{2}=\frac{1}{LC}=\frac{1}{2\times 18\times {10}^{-6}}``
    `` \Rightarrow {\omega }^{2}=\frac{{10}^{6}}{36}``
    `` \Rightarrow \omega =\frac{{10}^{3}}{6}``
    `` \Rightarrow 2\,\mathrm{\,\pi \,}f=\frac{{10}^{3}}{6}``
    `` \Rightarrow f=\frac{1000}{6\times 2\,\mathrm{\,\pi \,}}=26.539\,\mathrm{\,Hz\,}``
    `` =27\,\mathrm{\,Hz\,}``
  • #16-b
    What is the value of this maximum current?
    Ans : At resonance, reactance is minimum.
    Minimum Reactance, Z = R
    Maximum current (I) is given by,
    `` I=\frac{E}{R}``
    `` \Rightarrow I\mathit{=}\frac{20}{10\times {10}^{3}}``
    `` \Rightarrow I=\frac{2\,\mathrm{\,A\,}}{{10}^{3}}=2\,\mathrm{\,mA\,}``
    Page No 330: