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QJEEAdv21Pha612: JEE Advanced 2021 Physics - Class 12 JEEAdv21Pha612 Solution: In a circuit, a metal filament lamp is connected in series with …

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μ\mu F across a 200 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 the supply voltage is φ\varphi . Assume, π\pi 3\sqrt 3 ≈\approx 5.

The value of φ\varphi is ____________.

Step-by-Step Verified Solution

For lamp,

p=V2Rp = {{{V^2}} \over R}

R=100×100500=20ΩR = {{100 \times 100} \over {500}} = 20\Omega

i=VR=10020=5A⇒i=VR2+XC2i = {V \over R} = {{100} \over {20}} = 5A \Rightarrow i = {V \over {\sqrt {{R^2} + X_C^2} }}

⇒5=200(20)2+(12π×50×C)2\Rightarrow 5 = {{200} \over {\sqrt {{{(20)}^2} + {{\left( {{1 \over {2\pi \times 50 \times C}}} \right)}^2}} }}

⇒400+(1100πC)2=1600\Rightarrow 400 + {\left( {{1 \over {100\pi C}}} \right)^2} = 1600

⇒(1100πC)2=1200⇒1π2C2=1200×104\Rightarrow {\left( {{1 \over {100\pi C}}} \right)^2} = 1200 \Rightarrow {1 \over {{\pi ^2}{C^2}}} = 1200 \times {10^4}

⇒(πC)2=112×10−6\Rightarrow {(\pi C)^2} = {1 \over {12}} \times {10^{ - 6}}

⇒C2=112π2×10−6\Rightarrow {C^2} = {1 \over {12{\pi ^2}}} \times {10^{ - 6}}

⇒\Rightarrow C=112π×10−3=100×10−6F=100μFC = {1 \over {\sqrt {12} \pi}} \times {10^{ - 3}} = 100 \times {10^{ - 6}}F = 100\mu F

V=VC2+VR2V = \sqrt {V_C^2 + V_R^2}

200=VC2+1002200 = \sqrt {V_C^2 + {{100}^2}}

⇒VC=1003\Rightarrow {V_C} = 100\sqrt 3

tan⁡ϕ=XCR=VCVR=1003100=3\tan \phi = {{{X_C}} \over R} = {{{V_C}} \over {{V_R}}} = {{100\sqrt 3 } \over {100}} = \sqrt 3

ϕ=60∘\phi = 60^\circ

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