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An inductor 20 mH, a capacitor 100 microF and a
resistor 50 ohm are connected in series across a
source of emf, V = 10 sin 314 t. The power loss in
the circuit is
(1) 0·79 W
(2) 0·43 W
(3) 1·13 W
(4) 2·74 W

Shresthahahaha , 5 Years ago
Grade 12
anser 1 Answers
Arun

Last Activity: 5 Years ago

Dear student
 
Pav = (Vrms/Z)^2  R
 
Z = 56 ohm
 
hence Pav = (10/56sqrt(2))^2 * 50 = 0.79
 
Regards
Arun (askIITians forum expert)

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