In a series LCR circuit, the inductive reactance ( X L ) is 10 Ω and the capacitive reactance ( X C ) is 4 Ω. The resistance ( R ) in the circuit is 6 Ω. The power factor of the circuit is:

Question

In a series LCR circuit, the inductive reactance ( X L ) is 10 Ω and the capacitive reactance ( X C ) is 4 Ω. The resistance ( R ) in the circuit is 6 Ω. The power factor of the circuit is: (A) (1/2) (B) (1/2 √2) (C) (1/√2) (D) (√3/2)

Tardigrade Admin · Accepted Answer

We know that power factor is cos φ, cos φ=( R / Z ) dots(1) Z =√ R 2+( X L - X C )2 ldots(2) (ω L -1 / ω C )

⇒ Z=√62+(10-4)2 ⇒ Z=6 √2 mid cos φ=(6/6 √2) cos φ=(1/√2)

Q. In a series LCR circuit, the inductive reactance $(X_{L})$ is $10Ω$ and the capacitive reactance $(X_{C})$ is $4Ω$ . The resistance $(R)$ in the circuit is $6Ω$ . The power factor of the circuit is:

$\frac{1}{2}$

$\frac{1}{2 2}$

$\frac{1}{2}$

$\frac{3}{2}$

We know that power factor is $cos ϕ$ ,
$cos ϕ = \frac{R}{Z} \dots (1)$
$Z = R^{2} + (X_{L} - X_{C})^{2} \dots (2)$
$(ω L - 1/ ω C)$

$\Rightarrow Z = 6^{2} + (10 - 4)^{2}$
$\Rightarrow Z = 62 ∣ cos ϕ = \frac{6}{6 2}$
$cos ϕ = \frac{1}{2}$

Q. In a series LCR circuit, the inductive reactance (XL​) is 10Ω and the capacitive reactance (XC​) is 4Ω. The resistance (R) in the circuit is 6Ω. The power factor of the circuit is:

21​

22​1​

2​1​

23​​

We know that power factor is cosϕ, cosϕ=ZR​…(1) Z=R2+(XL​−XC​)2​…(2) (ωL−1/ωC) ⇒Z=62+(10−4)2​ ⇒Z=62​∣cosϕ=62​6​ cosϕ=2​1​

Q. In a series LCR circuit, the inductive reactance $(X_{L})$ is $10Ω$ and the capacitive reactance $(X_{C})$ is $4Ω$ . The resistance $(R)$ in the circuit is $6Ω$ . The power factor of the circuit is:

$\frac{1}{2}$

$\frac{1}{2 2}$

$\frac{1}{2}$

$\frac{3}{2}$

We know that power factor is $cos ϕ$ ,
$cos ϕ = \frac{R}{Z} \dots (1)$
$Z = R^{2} + (X_{L} - X_{C})^{2} \dots (2)$
$(ω L - 1/ ω C)$

$\Rightarrow Z = 6^{2} + (10 - 4)^{2}$
$\Rightarrow Z = 62 ∣ cos ϕ = \frac{6}{6 2}$
$cos ϕ = \frac{1}{2}$