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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?
$RC$ circuit with $R = 1\, k\,\Omega $ and $C = 1\, \mu F$
$RL$ circuit with $R = 1\, k\,\Omega $ and $L = 10\, mH$
$RL$ circuit with $R = 1\, k\,\Omega $ and $L = 1\, mH$
$RC$ circuit with $R = 1\, k\,\Omega $ and $C = 10\, \mu F$
Solution
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 $\mathrm{X}_{\mathrm{c}}=\frac{1}{10 \times 10^{-6} \times 100}=10^{3} \,\Omega$
