JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2019
- Hard
- MCQ
A circuit connected to an \(ac\) source of \(emf\) \(e = e_0\, sin\, (1000t)\) with \(t\) in seconds, gives a phase difference of \(\frac{\pi }{4}\) between the \(emf\) \(e\) and current \(i\). Which of the following circuits will exhibit this?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(RC\) circuit with \(R = 1\, k\,\Omega \) and \(C = 10\, \mu F\)
Step-by-step Solution
Detailed explanation
Given phase difference \(=\frac{\pi}{4}\) and \(\omega=100\) \(rad/s\)
\(\Rightarrow\) Reactance \((X)=\) Resistance \((R)\) Now by checking option.
Option \((A)\)
\(\mathrm{R}=1000 \,\Omega\) and \(\mathrm{X}_{\mathrm{c}}=\frac{1}{10^{-6} \times 100}=10^{4} \,\Omega\)
Option \((B)\)
\(\mathrm{R}=10^{3} \,\Omega\) and \(\mathrm{X}_{\mathrm{L}}=10 \times 10^{-3} \times 100=1 \,\Omega\)
Option \((\mathrm{C})\)
\(\mathrm{R}=10^{3} \,\Omega\) and \(\mathrm{X}_{\mathrm{L}}=10^{-3} 100=10^{-1}\, \Omega\)
Option \((D)\)
\(\mathrm{R}=10^{3} \,\Omega\) and \(\mathrm{X}_{\mathrm{c}}=\frac{1}{10 \times 10^{-6} \times 100}=10^{3} \,\Omega\)
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