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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2017
  • Hard
  • MCQ

A sinusoidal voltage of peak value \(283\, V\) and angular frequency \(320/s\) is applied to a series \(LCR\) circuit. Given that \(R\, = 5\,\Omega \) , \(L\,= 25\, mH\) and \(C\, = 1000\, \mu F\). The total impedance, and phase difference between the voltage across the source and the current will respectively be

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Answer & Solution

Correct Answer

(B) \(7\,\Omega \,\,\) and \(45^o\)

Step-by-step Solution

Detailed explanation

Given,

\(\mathrm{V}_{0}=283 \,\mathrm{volt}, \omega=320,\, \mathrm{R}=5 \,\Omega, \mathrm{L}=25 \,\mathrm{mH}, \mathrm{C} =1000 \,\mu \mathrm{F}\)

\(x_{L}=\omega L=320 \times 25 \times 10^{-3}=8\, \Omega\)

\(x_{C}=\frac{1}{\omega C}=\frac{1}{320 \times 1000 \times 10^{-6}}=3.1 \,\Omega\)

Total impedance of the circuit:

\(z=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}=\sqrt{25+(4.9)^{2}}=7\, \Omega\)

Phase difference between the voltage and current

\(\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}\)

\(\tan \phi=\frac{4.9}{5} \approx 1 \Rightarrow \phi=45 \%\)

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