You visited us 7 times! Enjoying our articles? Unlock Full Access!

Power Factor

A lamp consum...

Question

A lamp consumes only $50 %$ of peak power in an ac circuit. What is the phase difference between the applied voltage and the circuit current?

\frac{π}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{π}{3}$
$P = \frac{1}{2} v_{0} i_{0} c o s ψ$ ,

$P = P_{p e a k} . c o s ψ$

$\frac{1}{2} P_{p e a k} = P_{p e a k} c o s ψ = \frac{1}{2} = ψ = \frac{π}{3}$

Suggest Corrections

Similar questions

50

50 %

A C

V = 200 \sqrt{2} sin (ω t + \frac{π}{4})

i = - \sqrt{2} cos (ω t + \frac{π}{4})

A C

V = V_{0} sin 100 t

π / 4

LCR

AC

L

\frac{π}{3}

C

\frac{π}{3}

Join BYJU'S Learning Program