In an AC circuit, a resistance of R ohm is connected in series with an inductor of self inductance L. If phase angle between voltage and current be 45°, the value of inductive reactance (XL) will be equal to

Question

In an AC circuit, a resistance of R ohm is connected in series with an inductor of self inductance L. If phase angle between voltage and current be 45°, the value of inductive reactance (XL) will be equal to (A) R (B)  ( R/ 8)  (C)  ( R/ 4)  (D)  ( R/ 2)

Tardigrade Admin · Accepted Answer

The phasor diagram in L-R circuit is given below.

Given, the phase angle between voltage and current

ϕ = 4 5^{\circ}

So,

tan ϕ = \frac{V _{L}}{V _{R}} = \frac{I X _{L}}{I R}

Here,

X_{L}

= inductive reactance

\Rightarrow tan 4 5^{\circ} = \frac{X _{L}}{R}

\Rightarrow 1 = \frac{X _{L}}{R} \Rightarrow X_{L} = R

Q. In an AC circuit, a resistance of R ohm is connected in series with an inductor of self inductance L. If phase angle between voltage and current be $4 5^{\circ}$ , the value of inductive reactance $(X_{L})$ will be equal to

R

$\frac{R}{8}$

$\frac{R}{4}$

$\frac{R}{2}$

Q. In an AC circuit, a resistance of R ohm is connected in series with an inductor of self inductance L. If phase angle between voltage and current be 45∘, the value of inductive reactance (XL​) will be equal to

R

8R​

4R​

2R​

The phasor diagram in L-R circuit is given below. Given, the phase angle between voltage and current ϕ=45∘ So, tanϕ=VR​VL​​=IRIXL​​ Here, XL​ = inductive reactance ⇒tan45∘=RXL​​ ⇒1=RXL​​⇒XL​=R

Q. In an AC circuit, a resistance of R ohm is connected in series with an inductor of self inductance L. If phase angle between voltage and current be $4 5^{\circ}$ , the value of inductive reactance $(X_{L})$ will be equal to

$\frac{R}{8}$

$\frac{R}{4}$

$\frac{R}{2}$