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JEE Advanced 2021Paper 2
Exam
JEE Advanced
Year
2021
Paper
Paper 2
Subject
Physics
Chapter
Alternating Current

JEE Advanced 2021 · Paper 2 · Physics · Alternating Current

In a circuit, a metal filament lamp is connected in series with a...

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance CμF\mathrm{C} \mu F across a 200 V,50 Hz200 \mathrm{~V}, 50 \mathrm{~Hz} supply. The power consumed by the lamp is 500 W500 \mathrm{~W} while the voltage drop across it is 100 V100 \mathrm{~V}. Assume that there is no inductive load in the circuit. Take rmsr m s values of the voltages. The magnitude of the phaseangle (in degrees) between the current and the supply voltage is φ\varphi. Assume, π3≈5\pi \sqrt{3} \approx 5.
The value of ϕϕ is _____.In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μ F across a 200 ~V, 50In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μ F across a 200 ~V, 50

Correct answer

0

Step-by-step solution

Given,

P=500WP = 500W

VR=100V{V}_{R} = 100V

f=50Hzf = 50Hz

Vrms=200V{V}_{rms} = 200V

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μ F across a 200 ~V, 50

P=VR2R⇒R=VR2P⇒R=(100)2500P = \dfrac{{{V}_{R}}^{2}}{R} ⇒ R = \dfrac{{{V}_{R}}^{2}}{P} ⇒ R = \dfrac{{\left(100\right)}^{2}}{500}

∴R=20Ω∴ R = 20Ω

Now current in the circuit,

i=irms=voltageacrossresistorresistancei=10020i=5Ai = {i}_{rms} = \dfrac{voltageacrossresistor}{resistance}i = \dfrac{100}{20}i = 5A

Phasor diagram of the given circuit for voltage,

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μ F across a 200 ~V, 50

Now,

V2=VC2+VR2⇒2002=VC2+1002⇒VC=1003V{V}^{2} = {{V}_{C}}^{2} + {{V}_{R}}^{2} ⇒ {200}^{2} = {{V}_{C}}^{2} + {100}^{2} ⇒ {V}_{C} = 100\sqrt{3}V

Now,

VC=1003V=IXC⇒XC=10035=203Ω{V}_{C} = 100\sqrt{3}V = I{X}_{C} ⇒ {X}_{C} = \dfrac{100\sqrt{3}}{5} = 20\sqrt{3}Ω

and

XC=1(2\pif)C⇒203=1(2π×100)C⇒C=100μF{X}_{C} = \dfrac{1}{\left(2\pif\right)C} ⇒ 20\sqrt{3} = \dfrac{1}{\left(2\pi \times 100\right)C} ⇒ C = 100\mu F

From the phasor diagram,

tan⁡(ϕ)=XCR=20320=3⇒ϕ=60°\tan \left(ϕ\right) = \dfrac{{X}_{C}}{R} = \dfrac{20\sqrt{3}}{20} = \sqrt{3} ⇒ ϕ = 60 °

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