Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsEasy2 minPYQ_2020

PhysicsEasysingle choice

In a series $LCR$ circuit, the voltage drop across the resistance is $8 V$ , across the inductor is $6 V$ and across the capacitor is $12 V$ . Then,

Options:

Answer:

Solution:

Since, $\cos ϕ = \frac{R}{Z} = \frac{I R}{I Z} = \frac{8}{10} = \frac{4}{5}$

(Also $\cos ϕ$ can never be greater than $1$ )

Hence, (c) is wrong

Also $I X_{C} > I X_{L} \Rightarrow X_{C} > X_{L}$

$∴$ Current will be leading.

In an $LCR$ circuit,

$V = \sqrt{{(V_{L} - V_{C})}^{2} + V_{R}^{2}} = \sqrt{{(6 - 12)}^{2} + 8^{2}}$

$V = 10 V$ , which is less than the voltage drop across the capacitor.

Subject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: PYQ_2020

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