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  • Shown in the figure is an R-L circuit. Just after the Key (K) is closed <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/07/phy-q45-f21.PNG" style="height:121px; width:2

Shown in the figure is an R-L circuit. Just after the Key (K) is closed

Option: 1

 the current in the circuit is E/R


Option: 2

no potential drop across the resistor


Option: 3

potential drop across the inductor is E/2


Option: 4

heat is dissipated in the circuit in the resistance


Answers (1)

best_answer

The current in the circuit at the instant of closing the Key (K) is equal to io. Current at any time in transient state is given by

\mathrm{\mathrm{i}=\mathrm{i}_0\left(1-\mathrm{e}^{-\frac{\mathrm{t}}{\mathrm{\tau}}}\right) \text { where } \tau=\frac{\mathrm{L}}{\mathrm{R}}}

Putting t = 0, i = 0

Potential drop across resistor is zero

\mathrm{\Rightarrow} The voltage drop across the inductor is \mathrm{E^{\prime}} that oppose the applied emf E

\mathrm{\Rightarrow}\mathrm{E^{\prime}=E} numerically. Since the circuit current is zero, the heat loss (proportional to \mathrm{\mathrm{i}_0^2 \mathrm{R}}) will be zero

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Gunjita

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