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JEE Advanced Physics Alternating Current 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Physics Question (2021) — Solution

Question

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C 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 r m s values of the voltages. The magnitude of the phaseangle (in degrees) between the current and the supply voltage is . Assume, 3 5. The value of  ϕ  is _____.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Given,  P = 500   W V R = 100   V f = 50   Hz V r m s = 200   V P = V R 2 R ⇒ R = V R 2 P ⇒ R = 100 2 500 ∴     R = 20   Ω Now current in the circuit, i = i r m s = voltage   across   resistor resistance i = 100 20 i = 5   A Phasor diagram of the given circuit for voltage,   Now,  V 2 = V C 2 + V R 2 ⇒ 200 2 = V C 2 + 100 2 ⇒ V C = 100 3   V Now,  V C = 100 3   V = I X C ⇒ X C = 100 3 5 = 20 3   Ω and  X C = 1 2 π f C ⇒ 20 3 = 1 2 π × 100 C ⇒ C = 100   μF From the phasor diagram, tan ϕ = X C R = 20 3 20 = 3 ⇒ ϕ = 60 °

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