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-/1 points 31 OSUniPhys1 31.4. P.053. My For the circuit shown below, E = 14 V, L 4.4 mH, and R = 4.5 . After steady state is

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The answer for above problem is explained below.

Current through Inductor In a lev 4.52 - | In 2 3.11 A Current, iz In eft [4x16 4.52 x 4.4x10²4 = (3.1A) e 1 = (3.11A) *[0.66=> ✓ z [140] * [0.66425) =) Y = 19.2995 v Voltage across Resistor VR = IR = 2.0658AX 4.52 ve = 9.297V

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