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PhysicsEMI/ACSeries AC CircuitsHard2 minPYQ_2011

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A coil of resistance $10Ω$ and inductance $5 H$ is connected to a $100 V$ battery. Then the energy stored in the coil is

Answer:

Solution:

Final current, $I = \frac{E}{R} = 10 A$ Energystored in the magnetic field $U = \frac{1}{2} L i \frac{1}{2} \times 5 \times (10)^{2} = 250 J$

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

⏱ 2mℹ️ Source: PYQ_2011

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