Question:

Find the magnitudes of currents \( i_1 \), \( i_2 \), and \( i \) with the help of the given circuit, when (i) just at the moment switch S is pressed and (ii) S is pressed for a long time. 

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In an RL circuit, the inductor initially resists current flow and allows full current flow after a long time.
Updated On: Mar 4, 2025
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Solution and Explanation

(i) Just at the moment switch \( S \) is pressed: At \( t = 0 \), the inductor opposes any change in current, so the initial current through the inductor is zero. \[ i_2 = 0, \quad i_1 = \frac{10}{5} = 2 \text{ A} \] \[ \boxed{i = i_1 = 2 \text{ A}} \] (ii) After a long time: The inductor acts as a short circuit, and the total resistance becomes: \[ R_{\text{eq}} = 5 + 10 = 15 \, \Omega \] \[ i = \frac{10}{15} = \frac{2}{3} \text{ A} \] Current division: \[ i_1 = \frac{2}{3} \times \frac{10}{15} = \frac{2}{3} \text{ A}, \quad i_2 = \frac{2}{3} \times \frac{5}{15} = \frac{1}{3} \text{ A} \] \[ \boxed{i_1 = \frac{2}{3} \text{ A}, \quad i_2 = \frac{1}{3} \text{ A}} \]
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