An inductor 20 mH, a capacitor 50 and a resistor 40 are connected in series across a source of emf The power loss in A.C. circuit is [2016]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
An inductor 20 mH, a capacitor 50 $μ F$ and a resistor 40 $Ω$ are connected in series across a source of emf $V = 10 \sin 340 t .$ The power loss in A.C. circuit is [2016]

1 0.76 W

2 0.89 W

3 0.51 W

4 0.67 W

Ans.
(3)

Here, $L = 20$ $mH = 20 \times 10^{- 3}$ $H,$

$C = 50$ $μ F = 50 \times 10^{- 6}$ $F,$ $R = 40$ $Ω$

$V = 10 \sin 340 t = V_{0} \sin ω t$

$ω = 340 rad$ $s^{- 1}, V_{0} = 10$ $V$

$X_{L} = ω L = 340 \times 20 \times 10^{- 3} = 6.8$ $Ω$

$X_{C} = \frac{1}{ω C} = \frac{1}{340 \times 50 \times 10^{- 6}} = \frac{10^{4}}{34 \times 5} = 58.82$ $Ω$

$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}} = \sqrt{{(40)}^{2} + {(58.82 - 6.8)}^{2}}$

$= \sqrt{{(40)}^{2} + {(52.02)}^{2}} = 65.62$ $Ω$

The peak current in the circuit is

$I_{0} = \frac{V_{0}}{Z} = \frac{10}{65.62} A, \cos ϕ = \frac{R}{Z} = (\frac{40}{65.62})$

Power loss in A.C. circuit, $= V_{rms} I_{rms} \cos ϕ = \frac{1}{2} V_{0} I_{0} \cos ϕ$

$= \frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62} = 0.46$ $W$

Want Us to Email you About Special Offers & Updates?