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In the circuit shown the switch S is shifted to position 2 from position 1 at t = 0, having been in position 1 for a
long time. The current in the circuit just after shifting of the switch will be (battery and both the inductors are ideal)

 

Option: 1

\mathrm{\frac{4}{5} \cdot \frac{\varepsilon}{\mathrm{R}}}


Option: 2

\mathrm{\frac{5}{4} \cdot \frac{\varepsilon}{\mathrm{R}}}


Option: 3

\mathrm{\frac{5}{9} \frac{\varepsilon}{\mathrm{R}}}


Option: 4

\mathrm{\frac{\varepsilon}{\mathrm{R}}.}


Answers (1)

best_answer

Let \mathrm{i_1} is the current in circuit before shifting

\mathrm{i_1=\frac{\varepsilon}{R}}

Since the flux associated with the inductors will be same just before and just after shifting 

\mathrm{\begin{aligned} & \therefore i_1 \cdot 5 \mathrm{~L}=\mathrm{i}_2 (9 \mathrm{~L}) \\ & \mathrm{i}_2=\frac{5}{9} \frac{\varepsilon}{\mathrm{R}} . \end{aligned}}

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Sayak

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