A series resonant LCR circuit has a quality factor (Q-factor) 0.4. If R = 2 kΩ, C = 0.1 μ F, then the value of inductance is

Question

A series resonant LCR circuit has a quality factor (Q-factor) 0.4. If R = 2 kΩ, C = 0.1 μ F, then the value of inductance is (A) 0.1 H (B) 0.064 H (C) 2 H (D) 5 H

Tardigrade Admin · Accepted Answer

Q = (1/R) √(L/C) or (L/C) = (QR)2 ∴ : :: L = (0.4 × 2 × 103)2 × 0.1 × 10-6 = 0.064 H

Q. A series resonant $L CR$ circuit has a quality factor ( $Q$ -factor) 0.4. If $R = 2 k Ω, C = 0.1 μ F$ , then the value of inductance is

0.1 H

0.064 H

2 H

5 H

$Q = \frac{1}{R} \frac{L}{C}$ or $\frac{L}{C} = (QR)^{2}$
$∴ L = (0.4 \times 2 \times 1 0^{3})^{2} \times 0.1 \times 1 0^{- 6} = 0.064 H$

Q. A series resonant LCR circuit has a quality factor (Q-factor) 0.4. If R=2kΩ,C=0.1μF, then the value of inductance is

0.1 H

0.064 H

2 H

5 H

Q=R1​CL​​ or CL​=(QR)2 ∴L=(0.4×2×103)2×0.1×10−6=0.064H

Q. A series resonant $L CR$ circuit has a quality factor ( $Q$ -factor) 0.4. If $R = 2 k Ω, C = 0.1 μ F$ , then the value of inductance is

$Q = \frac{1}{R} \frac{L}{C}$ or $\frac{L}{C} = (QR)^{2}$
$∴ L = (0.4 \times 2 \times 1 0^{3})^{2} \times 0.1 \times 1 0^{- 6} = 0.064 H$