JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Alternating Current

A 0.07 H inductor and a

12Ω

resistor are connected in series to a

220 V

50 Hz

ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take

π

\frac{22}{7}

]

Answer: (B)

Solution

The inductive reactance at

50

Hz is

X_{L} = 2 π f L = 2 \times \frac{22}{7} \times 50 \times 0.07 = 22 Ω

so the impedance is

Z = R^{2} + X_{L}^{2} = 1 2^{2} + 2 2^{2} = 628 \approx 25 Ω

and the current is

I = \frac{V}{Z} = \frac{220}{25} = 8.8

A. In a series inductive circuit the current lags the applied voltage by

ϕ = tan^{- 1} \frac{X _{L}}{R} = tan^{- 1} \frac{22}{12}

. The circuit carries

8.8

A with a phase angle

tan^{- 1} (\frac{11}{6})

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option