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
Pick an option to check your answer.
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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