In the L - R circuit shown in the figure- the key K is closed at time t -0-A- the current through R1 decreases with time- tB- the voltage difference across L increases with tC- the magnetic energy stored- in the steady state- equals 1-2 L E 2- R 22D- the total current through the battery is E - R 1- E - R 2- in steady state

The correct options are C the magnetic energy stored, in the steady state, equals 1/2 L E^2/(R_2)^2 D the total current through the battery is E/R_1 ...

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Byju's Answer

Standard XII

Physics

Alternating Current Applied to an Inductor

In the L - R ...

Question

In the L – R circuit shown in the figure, the key K is closed at time t = 0.

the current through R1 decreases with time, t

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the voltage difference across L increases with t

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the magnetic energy stored, in the steady state, equals $\frac{1}{2} L \frac{E^{2}}{(R_{2})^{2}}$

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the total current through the battery is $\frac{E}{R_{1}} + \frac{E}{R_{2}}$ , in steady state

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Solution

The correct options are
C
the magnetic energy stored, in the steady state, equals $\frac{1}{2} L \frac{E^{2}}{(R_{2})^{2}}$

D
the total current through the battery is $\frac{E}{R_{1}} + \frac{E}{R_{2}}$ , in steady state

The current through R1 is constant i.e., $i_{1} = \frac{E}{R_{1}}$ , current through $R_{2}$ varies. At steady state the inductor will get short circuited, so the current through the battery will be $\frac{E}{R_{1}} + \frac{E}{R_{2}}$

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