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PhysicsEMI/ACSeries AC CircuitsMedium2 minPYQ_2020

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The current through an inductor of impedance $10 Ω$ lags behind the voltage by a phase of $60 °$ when just the inductor is connected to the ac source. Now the inductor is connected to a $5 Ω$ resistance in series, then the net impedance of the circuit is

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Answer:

Solution:

$10 = \sqrt{r^{2} + X_{L}^{2}}$ and $\frac{X_{L}}{r} = \tan 60 °$

$10 = \sqrt{r^{2} + {(r \sqrt{3})}^{2}}$ or $r = 5 Ω, X_{L} = 5 \sqrt{3} Ω$

$Z = \sqrt{{(5 + 5)}^{2} + {(5 \sqrt{3})}^{2}} = \sqrt{175} = 13.2 Ω$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: PYQ_2020

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