Physics · Alternating Current

JEE Advanced 2021 — Paper 2 — Question 30

In the circuit, a metal filament lamp is connected in series with a capacitor of capacitance CμF\mathrm{C} \mu \mathrm{F} across a 200 V,50200 \mathrm{~V}, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V . Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and supply voltage is ϕ\phi. Assume, π3≈5\pi \sqrt{3} \approx 5.

The value of ϕ\phi is   ⁣ ⁣  ⁣ ⁣ \text{ }\!\!~\!\!\text{ } .

Answer: 60.00

Numerical answer — enter this value.

Step-by-step solution

cosϕ=RZ=2040=12\text{cos}\phi =\frac{\text{R}}{\text{Z}}=\frac{20}{40}=\frac{1}{2} ⇒ϕ=60∘\Rightarrow \phi ={{60}^{\circ }}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 2
Subject
Physics
Chapter
Alternating Current
Topic
Series L-R, R-C, L-C Circuits with AC Source
In the circuit, a metal filament lamp is connected in series with a… | JEE Advanced 2021 PYQ with Solution · DhiX AI