JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Alternating Current

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance

C μ 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

. Assume,

π 3 \approx 5

.

The value of C is ____.

Answer: 100

Solution

Given:

ε_{rms} = 200

V_{R} = 100

P = 500

f = 50

Hz

Using Pythagoras theorem for series RC circuit:

V_{C}^{2} + V_{R}^{2} = ε_{rms}

V_{C}^{2} + 10 0^{2} = 20 0^{2}

V_{C} = 40000 - 10000 = 30000 = 1003

V

Phase angle:

tan ϕ = \frac{V _{C}}{V _{R}} = \frac{100 3}{100} = 3

ϕ = 6 0^{\circ}

Power:

P = ε_{rms} I_{rms} cos ϕ

500 = 200 \times I_{rms} \times \frac{1}{2}

I_{rms} = 5

A

Resistance:

R = \frac{V _{R}}{I _{rms}} = \frac{100}{5} = 20

Ω

Impedance:

Z = \frac{ε _{rms}}{I _{rms}} = \frac{200}{5} = 40

Ω

Capacitive reactance:

X_{C} = Z^{2} - R^{2} = 1600 - 400 = 1200 = 203

Ω

Capacitance:

C = \frac{1}{ω X _{C}} = \frac{1}{2 π f X _{C}} = \frac{1}{2 π \times 50 \times 20 3}

Using

π 3 \approx 5

C = \frac{1}{2 \times 5 \times 50 \times 20} = \frac{1}{10000} = 1 0^{- 4}

= 100

μF

So

C = 100.00

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Numerical answer

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Alternating Current

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