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Byju's Answer

Standard XII

Physics

Impedance in RLC Circuit

A pure resist...

Question

A pure resistive circuit element X when connected to an AC supply of peak voltage 200 V gives a peak current of 5 V which is in phase with voltage. A second circuit elements Y. when connected to the same AC supply also gives the of peak current but the lags behind by $90^{\circ}$ . if the series combination of X and Y is connected to the same , what will be the ms value of current?

\frac{40}{\sqrt{2}} a m p

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40 \sqrt{2} a m p

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\frac{10}{5} a m p

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5 amp

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Solution

The correct option is B $40 \sqrt{2} a m p$
$R = \frac{E_{0}}{l_{0}} = \frac{200}{5} = 40 Ω$
we know
$X_{L} = \frac{E_{0}}{l_{0}} = \frac{200}{5} = 40 Ω$
Total impedence
$z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{40^{2} + 40^{2}}$
$= 40 \sqrt{2} Ω$

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The correct option is B 40√2ampR=E0l0=2005=40Ωwe knowXL=E0l0=2005=40ΩTotal impedence z=√R2+X2L=√402+402=40√2Ω

The correct option is B $40 \sqrt{2} a m p$
$R = \frac{E_{0}}{l_{0}} = \frac{200}{5} = 40 Ω$
we know
$X_{L} = \frac{E_{0}}{l_{0}} = \frac{200}{5} = 40 Ω$
Total impedence
$z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{40^{2} + 40^{2}}$
$= 40 \sqrt{2} Ω$