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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Alternating Current

BITSAT 2016 · Physics · Alternating Current

A resistor of resistance R , capacitor of capacitance C and...

A resistor of resistance RR, capacitor of capacitance CC and inductor of inductance LL are connected in parallel to AC power source of voltage ε0sin⁡ωt\varepsilon_{0} \sin \omega t. The maximum current through the resistance is half of the maximum current through the power source. Then value of R\mathrm{R} isA resistor of resistance R , capacitor of capacitance C and inductor of inductance L are connected in parallel to AC

Options

  1. A3∣ωC−1ωL∣\frac{\sqrt{3}}{\left|\omega C-\frac{1}{\omega L}\right|}
  2. B3∣1ωC−ωL∣\sqrt{3}\left|\frac{1}{\omega C}-\omega L\right|
  3. C5∣1ωC−ωL∣\sqrt{5}\left|\frac{1}{\omega C}-\omega L\right|
  4. DNone of these

Correct answer

A. 3∣ωC−1ωL∣\frac{\sqrt{3}}{\left|\omega C-\frac{1}{\omega L}\right|}

Step-by-step solution

iR0(iR0)2+(ic0−iL0)2=12⇒ε0/R(ε0/R)2+(ε0ωC−ε0ωL)2=12⇒R=3(ωC−1ωL)\frac{i_{R_{0}}}{\sqrt{\left(i_{R_{0}}\right)^{2}+\left(i_{c_{0}}-i_{L_{0}}\right)^{2}}}=\frac{1}{2}\Rightarrow \frac{\varepsilon_{0} / R}{\sqrt{\left(\varepsilon_{0} / R\right)^{2}+\left(\varepsilon_{0} \omega C-\frac{\varepsilon_{0}}{\omega L}\right)^{2}}}=\frac{1}{2}\Rightarrow R=\frac{\sqrt{3}}{\left(\omega C-\frac{1}{\omega L}\right)} A resistor of resistance R , capacitor of capacitance C and inductor of inductance L are connected in parallel to AC
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