Explanation Given: The inductance of the coilis 10 mH . The value of the current at t = 0 − s is 4 mA . Concept used: Write the expression for the current through the inductor is i L ( t ) = 1 L ∫ t 1 t 2 v L ( t ) d t + i L ( t 1 ) ...... (1) Here, L is the inductance of the coil, i L ( t ) is the current through the inductor, i L ( t 1 ) is the initial current and v L ( t ) is the voltage across the inductor. Calculation: For time interval 0 ≤ t ≤ 5 μ s the voltage across inductor is 0 V . Substitute 10 mH for L , 0 V for v L ( t ) , 4 mA for i L ( t 1 ) , 0 μ s for t 1 and t for t 2 in equation (1). i L ( t ) = 1 10 mH ∫ 0 μ s t ( 0 V ) d t + 4 mA = 0 + 4 mA = 4 mA ( 1 A 1000 mA ) = 4 × 10 − 3 A For time interval 5 ≤ t ≤ 10 μ s the voltage across inductor is − 10 V . Substitute 10 mH for L , − 10 V for v L ( t ) , 4 mA for i L ( t 1 ) , 5 μ s for t 1 and t for t 2 in equation (1). i L ( t ) = 1 10 mH ( 1 H 1000 mH ) ∫ 5 μ s t ( − 10 V ) d t + 4 mA ( 1 A 1000 mA ) = − 10 V 10 × 10 − 3 H ∫ 5 μ s t d t + 4 × 10 − 3 A = − 1 V 1 × 10 − 3 H ( t − 5 μ s ) ( 1 s 10 6 μ s ) + 4 × 10 − 3 A = ( − 10 − 3 t + 5 × 10 − 3 + 4 × 10 − 3 ) A Simplify further. i L ( t ) = − 10 − 3 t + 9 × 10 − 3 A Substitute 10 μ s for t in above equation. i L ( t ) = − 10 − 3 ( 10 ) + 9 × 10 − 3 A = − 1 × 10 − 3 A For time interval 10 ≤ t ≤ 12 μ s the voltage across inductor is − 25 V . Substitute 10 mH for L , − 25 V for v L ( t ) , − 1 × 10 − 3 A for i L ( t 1 ) , 10 μ s for t 1 and t for t 2 in equation (1). i L ( t ) = 1 10 mH ( 1 H 1000 mH ) ∫ 10 μ s t ( − 25 V ) d t + ( − 1 × 10 − 3 A ) = − 25 V 10 × 10 − 3 H ∫ 10 μ s t d t − 1 × 10 − 3 A = − 25 V 10 × 10 − 3 H ( t − 10 μ s ) ( 1 s 10 6 μ s ) − 1 × 10 − 3 A = − 2

13th Edition

ISBN: 9780133923919

Author: Boylestad, Robert L.

Publisher: Pearson Education

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Chapter 11, Problem 36P

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Introductory Circuit Analysis

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