Question

An alternating e.m.f. of frequency 50 Hz is applied across series combination of resistor 'R' and inductance `(sqrt3/pi)`H. If the phase difference between applied e.m.f. and current 'I' is 60°, then the value of R is `[tan 60^circ = sqrt3, sin 60^circ = sqrt3/2, cos 60^circ = 1/2]`

Options

• 300 Ω

• 50 Ω

• 200 Ω

• 100 Ω

Answer 1

100 Ω

Explanation:

L = `sqrt3/pi`H, f = 50 Hz, Φ = 60°

X_L = 2πfL = 2π × 50 × `sqrt3/pi = 100sqrt3 omega`

Z = `sqrt("R"^2 + "X"_"L"^2)`

∴ cos 60° = `1/2 = "R"/sqrt("R"^2 + "X"_"L"^2)`

∴ `1/4 = "R"^2/("R"^2 + "X"_"L"^2)`

∴ `4"R"^2 = "R"^2 + "X"_"L"^2`

∴ `3"R"^2 = "X"_"L"^2`

∴ `"R"^2 = "X"_"L"^2/3`

∴ R = `"X"_"L"/sqrt3 = (100 sqrt3)/sqrt3 = 100 Omega`