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We have three devices, a capacitor of $C = 1 μF$ , an inductor of L = 10mH and a resistor of ‘R’ $Ω$ when C and R are used in series combination and an alternating source of emf is connected across it, current leads the supply voltage by $45^{0}$ when L and R are used in series combination and the same supply voltage is applied across the combination, current lags the supply voltage by $45^{0}$ . Then R is

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$10 Ω$

$100 Ω$

$10 \sqrt{10} Ω$

$100 \sqrt{10} Ω$

answer is B.

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Detailed Solution

$\frac{\frac{1}{ωc}}{R} = \tan 45^{0} \Rightarrow ωRC = 1 - - - - (1) \frac{ωL}{R} = \tan 45^{0} \Rightarrow ωL = R_____(2) Solving, R = \sqrt{\frac{L}{C}} = 100 Ω$

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