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Let us try as your professor suggested as lim R->0 the equation becomes 0/0 which is undetrmied so let's use L'Hospital's Rule. we get $I(t)=\frac{V_b }L .t e^{-tR/L}=\frac{V_b.t}L$. Yes because an ideal inductor is short circuit for steady state. as you might know inductive reactense $X_L=j.\omega.L$ as it's DC, frequencie is zero it's practically a short circuit .since it violates kirchhoff's voltage law hence a large current flows in the circuit  .

Let us try as your professor suggested as lim R->0 the equation becomes 0/0 which is undetrmied so let's use L'Hospital's Rule. we get $$I(t)=\frac{V_b }L .t e^{-tR/L}=\frac{V_b.t}L \, .$$ Yes because an ideal inductor is short circuit for steady state. as you might know inductive reactance $X_L=j \omega L$ as it's DC, frequency is zero it's practically a short circuit since it violates Kirchhoff's voltage law hence a large current flows in the circuit.

Let us try as your professor suggested as lim R->0 the equation becomes 0/0 which is undetrmied so let's use L'Hospital's Rule. we get $I(t)=\frac{V_b }L .t e^{-tR/L}=\frac{V_b.t}L$.

Let us try as your professor suggested as lim R->0 the equation becomes 0/0 which is undetrmied so let's use L'Hospital's Rule. we get $I(t)=\frac{V_b }L .t e^{-tR/L}=\frac{V_b.t}L$. Yes because an ideal inductor is short circuit for steady state. as you might know inductive reactense $X_L=j.\omega.L$ as it's DC, frequencie is zero it's practically a short circuit .since it violates kirchhoff's voltage law hence a large current flows in the circuit .

Let us try as your professor suggested as lim R->0 the equation becomes 0/0 which is undetrmied so let's use L'Hospital's Rule. we get $I(t)=\frac{V_b .R}L .t e^{-tR/L}$. So the current in the Inductor decays exponentially.

Let us try as your professor suggested as lim R->0 the equation becomes 0/0 which is undetrmied so let's use L'Hospital's Rule. we get $I(t)=\frac{V_b }L .t e^{-tR/L}=\frac{V_b.t}L$.