A resistance of 40 Ω and an inductance of 95.5 mH are connected in series in a 50 cycle/second ac circuit. The impedance of this combination is very nearly:

Question

A resistance of 40 Ω and an inductance of 95.5 mH are connected in series in a 50 cycle/second ac circuit. The impedance of this combination is very nearly: (A) 30 Ω (B) 40 Ω (C) 50 Ω (D) 60 Ω

Tardigrade Admin · Accepted Answer

Z=√R2+XL2=√R2+(2 π v L)2 Z=√(40)2+4 π2 ×(50)2 ×(95.5 × 10-3)2 =50 Ω

Q. A resistance of $40Ω$ and an inductance of $95.5 m H$ are connected in series in a 50 cycle/second ac circuit. The impedance of this combination is very nearly:

$30Ω$

$40Ω$

$50Ω$

$60Ω$

$Z = R^{2} + X_{L}^{2} = R^{2} + (2 π vL)^{2}$
$Z = (40)^{2} + 4 π^{2} \times (50)^{2} \times (95.5 \times 1 0^{- 3})^{2}$
$= 50Ω$

Q. A resistance of 40Ω and an inductance of 95.5mH are connected in series in a 50 cycle/second ac circuit. The impedance of this combination is very nearly:

30Ω

40Ω

50Ω

60Ω