In the circuit shown in the figure, battery electromotive force E = 20 V. resistances Re=...

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In the circuit shown in the figure, battery electromotive force E = 20 V. resistances Re= 2 ohms, R₁ = 12ohms, R₂ = 4 ohms, and inductance L = 0.2 H. Long time after the switch S is closed find: 13. the current through the resistor R₂. (a) 5 A (b) 2 A (c) 4A (d) 3 A (e) 1 A 14. the potential difference VaVa-Vi across the inductor L. (a) 4 V (b) 20 V (e) 0 V (d) 2 V (e) 12 V 15. the energy stored in (a) 0.2 J (b) 3 J the inductor L.. (c) 144 J (d) 1.6 J (e) 0.9 J S O R₂ & R₁ ww Vb In the circuit shown in the figure, battery electromotive force E = 20 V. resistances Re= 2 ohms, R₁ = 12ohms, R₂ = 4 ohms, and inductance L = 0.2 H. Long time after the switch S is closed find: 13. the current through the resistor R₂. (a) 5 A (b) 2 A (c) 4A (d) 3 A (e) 1 A 14. the potential difference VaVa-Vi across the inductor L. (a) 4 V (b) 20 V (e) 0 V (d) 2 V (e) 12 V 15. the energy stored in (a) 0.2 J (b) 3 J the inductor L.. (c) 144 J (d) 1.6 J (e) 0.9 J S O R₂ & R₁ ww Vb

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