Solveeit Logo

Question

Physics Question on Alternating current

A lamp consumes only 25%25\% of the peak power in an ac circuit. The phase difference between the applied voltage and the current is

A

π6\frac{\pi}{6}

B

π3\frac{\pi}{3}

C

π4\frac{\pi}{4}

D

π2\frac{\pi}{2}

Answer

π3\frac{\pi}{3}

Explanation

Solution

According to the equation,
P=25100E0I0P =\frac{25}{100} E_{0} I_{0}
P=14E0I0=E0I02×2P =\frac{1}{4} E_{0} I_{0}=\frac{E_{0} I_{0}}{2 \times 2} ...(i)
We know that, P=E0I02cosϕP=\frac{E_{0} I_{0}}{2} \cos \phi ...(ii)
On comparing Eqs. (i) and (ii), we get
cosϕ=12\cos \phi =\frac{1}{2}
ϕ=cos1(12)\phi=\cos ^{-1}\left(\frac{1}{2}\right)
ϕ=π3\Rightarrow \phi =\frac{\pi}{3} or 6060^{\circ}