In a series circuit R -300 - L -0-9 H - C -2-0 - F and -1000 rad - sec- The impendence of the circuit isA- 400 -B- 1300 -C- 900 -D- 500 -

The correct option is D 500 Ω=√(R^2+[X_L X_c]^2)= √(R^2+ ( Lw 1/wc )^2) =√(300^2+ [ 900 1/10^3 × 2 × 10^ 6 ]^2) =√(300^2+[900 500]^2) = √(300^2+400^2) = 500 ...

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Standard XII

Physics

Impedance in RLC Circuit

In a series c...

Question

In a series circuit R = 300 $Ω$ , L = 0.9 H, C = 2.0 $μ$ F and $ω$ = 1000 rad/sec. The impendence of the circuit is

1300 Ω

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900 Ω

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500 Ω

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400 Ω

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Solution

The correct option is C $500 Ω$
$= \sqrt{R^{2} + [X_{L} - X_{c}]^{2}} = \sqrt{R^{2} + {(L w - \frac{1}{w c})}^{2}}$
$= \sqrt{300^{2} + {[900 - \frac{1}{10^{3} \times 2 \times 10^{- 6}}]}^{2}}$
$= \sqrt{300^{2} + [900 - 500]^{2}}$
$= \sqrt{300^{2} + 400^{2}}$
$= 500 Ω$

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In a series circuit R = 300 Ω , L = 0.9 H, C = 2.0 μF and ω = 1000 rad/sec. The impendence of the circuit is

The correct option is C 500 Ω=√R2+[XL−Xc]2=√R2+(Lw−1wc)2 =√3002+[900−1103×2×10−6]2 =√3002+[900−500]2 =√3002+4002 =500Ω

In a series circuit R = 300 $Ω$ , L = 0.9 H, C = 2.0 $μ$ F and $ω$ = 1000 rad/sec. The impendence of the circuit is

The correct option is C $500 Ω$
$= \sqrt{R^{2} + [X_{L} - X_{c}]^{2}} = \sqrt{R^{2} + {(L w - \frac{1}{w c})}^{2}}$
$= \sqrt{300^{2} + {[900 - \frac{1}{10^{3} \times 2 \times 10^{- 6}}]}^{2}}$
$= \sqrt{300^{2} + [900 - 500]^{2}}$
$= \sqrt{300^{2} + 400^{2}}$
$= 500 Ω$