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Question: A sinusoidal voltage of peak value 293 V and frequency 50 Hz is applied to a series LCR circuit in w...

A sinusoidal voltage of peak value 293 V and frequency 50 Hz is applied to a series LCR circuit in which R = 6 Ω\Omega, L = 25 mH and C=750μFC = 750\mu F The impedance of the circuit is

A

7.0 Ω\text{7.0 }\Omega

B

8.9 Ω\text{8.9 }\Omega

C

9.9 Ω\text{9.9 }\Omega

D

10.0 Ω\text{10.0 }\Omega

Answer

7.0 Ω\text{7.0 }\Omega

Explanation

Solution

: Here,

R=6ω,L=25mH=25×10−3H,R = 6\omega,L = 25mH = 25 \times 10^{- 3}H,

C=750μF=750×10−6F,υ=50HzC = 750\mu F = 750 \times 10^{- 6}F,\upsilon = 50HzXC=12πυC=12×3.14×50×750×10−6=4.25Ω∴XL=XC=7.85−4.25=3.6ΩX_{C} = \frac{1}{2\pi\upsilon C} = \frac{1}{2 \times 3.14 \times 50 \times 750 \times 10^{- 6}} = 4.25\Omega\therefore X_{L} = X_{C} = 7.85 - 4.25 = 3.6\Omega

Impedence of the series LCR circuit is

Z=R2+(XL−XC)2Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

∴Z=(6)2+(3.6)2\therefore Z = \sqrt{(6)^{2} + (3.6)^{2}}

=36+12.96=7.0Ω= \sqrt{36 + 12.96} = 7.0\Omega