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

Question

A lamp consumes only 50 % of peak power in an a.c. circuit. What is the phase difference between the applied voltage and the circuit current? (A) (π/6) (B) (π/3) (C) (π/4) (D) (π/2)

Tardigrade Admin · Accepted Answer

P=(1/2) Vo io cos φ ⇒ P=P text peak ⋅ cos φ ⇒ (1/2)(P text peak )=P text peak cos φ ⇒ cos φ=(1/2) ⇒ φ=(π/3)

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

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{2}$

$P = \frac{1}{2} V_{o} i_{o} cos ϕ$
$\Rightarrow P = P_{peak} \cdot cos ϕ$
$\Rightarrow \frac{1}{2} (P_{peak}) = P_{peak} cos ϕ$
$\Rightarrow cos ϕ = \frac{1}{2} \Rightarrow ϕ = \frac{π}{3}$

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

6π​

3π​

4π​

2π​