The current through an inductor of impedance 10 Ω lags behind the voltage by a phase of 60° when just the inductor is connected to the AC source. Now the inductor is connected to a 5Ω resistance in series, then the net impedance of the circuit is

Question

The current through an inductor of impedance 10 Ω lags behind the voltage by a phase of 60° when just the inductor is connected to the AC source. Now the inductor is connected to a 5Ω resistance in series, then the net impedance of the circuit is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-psqgudbqjjuv6vri.jpg" /> (A) 15 Ω (B) 12 Ω (C) 13.2 Ω (D) 18 Ω

Tardigrade Admin · Accepted Answer

10=√r2 + XL2 and (XL/r)=tan 60° 10=√r2 + (r √3)2 or r=5Ω, XL=5√3Ω Z=√(5 + 5)2 + (5 √3)2=√175=13.2 Ω

Q. The current through an inductor of impedance $10 Ω$ lags behind the voltage by a phase of $6 0^{\circ}$ when just the inductor is connected to the AC source. Now the inductor is connected to a $5Ω$ resistance in series, then the net impedance of the circuit is

$15 Ω$

$12 Ω$

$13.2 Ω$

$18 Ω$

$10 = r^{2} + X_{L}^{2}$ and $\frac{X _{L}}{r} = t an 6 0^{\circ}$
$10 = r^{2} + (r 3)^{2}$ or $r = 5Ω, X_{L} = 53 Ω$
$Z = (5 + 5)^{2} + (53)^{2} = 175 = 13.2 Ω$

Q. The current through an inductor of impedance 10Ω lags behind the voltage by a phase of 60∘ when just the inductor is connected to the AC source. Now the inductor is connected to a 5Ω resistance in series, then the net impedance of the circuit is

15Ω

12Ω

13.2Ω

18Ω