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Q.

We have three devices, a capacitor of C = 1μF , 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 450 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  450. Then R is

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a

10Ω

b

100Ω

c

1010Ω

d

10010 Ω

answer is B.

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Detailed Solution

1ωcR = tan450   ⇒ ωRC = 1−−−−(1) ωLR = tan450  ⇒ ωL = R_____(2) Solving, R = LC = 100Ω
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