A capacitor of reactance and a resistor of resistance are connected in series with an AC source of peak value . The power dissipation in the circuit is ______ W.

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.
A capacitor of reactance $4 \sqrt{3} Ω$ and a resistor of resistance $4 Ω$ are connected in series with an AC source of peak value $8 \sqrt{2} V$ . The power dissipation in the circuit is ______ W. [2024]

Ans.
(4)

$Z = \sqrt{R^{2} + {(X_{L})}^{2}} = \sqrt{4^{2} + {(4 \sqrt{3})}^{2}} = 8 Ω$

$V_{r m s} = \frac{V}{\sqrt{2}} = \frac{8 \sqrt{2}}{\sqrt{2}} = 8 V$

$I_{r m s} = \frac{V_{r m s}}{Z} = \frac{8}{8} = 1 A$

$Power dissipation P = V_{r m s} I_{r m s} \cos ϕ$

$\cos ϕ = \frac{R}{Z} = \frac{4}{\sqrt{{(4 \sqrt{3})}^{2} + 4^{2}}} = \frac{4}{\sqrt{48 + 16}} = \frac{4}{8} = \frac{1}{2}$

$∴$ $P = 8 \times 1 \times \frac{1}{2} = 4 W$

Want Us to Email you About Special Offers & Updates?