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An inductor of inductance L=400mH and re...

An inductor of inductance `L=400mH` and resistors of resistances `R_(1)=2 Omega` and `R_(2)=2 Omega` are connected to a battery of `emf 12 V` as shown in figure.The internal resistance of the battery is negligible.The switch `S` is closed at `t=0`.The potential drop across `L` as a function of time is:

Text Solution

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`I_1=E/R_1=(12)/2=6A ,E=L (dl_2)/(dt)+R_2xxl_2`
`I_2=I_0(1-e^(-t//t_c)), rArr I_0=E/R_2=(12)/2=6A`
`t_c=L/R=(400xx10^(-3))/2=0.2,I_2=6(1-e^(-t//0.2))`
Potential drop across L
`V_L=E-R_2I_2=12-2xx6(1-e^(-bt)),=12e^(-5t)`
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