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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsEasy2 minPYQ_2020

PhysicsEasysingle choice

A coil has self inductance $L = 0.04 H$ and resistance $R = 12 Ω$ . When it is connected to $220 V$ , $50 Hz$ supply; what will be the current flowing through the coil?

Options:

Answer:

Solution:

$L = 0.04 H$

$R = 12 Ω$

$V = 220 V, 50 Hz$

Impedance $z = \sqrt{R^{2} + w^{2} z^{2}}$

$w = 2 π f = 2 π \times 50 = 100 π$

$z = \sqrt{144 + {(100 π)}^{2} \times {(0.04)}^{2}}$

$= 17.38$

$i = \frac{v}{z} = \frac{220}{17.38} = 12.7 A .$

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

⏱ 2mℹ️ Source: PYQ_2020

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