A.An LR circuit is hooked up to a battery as shown in the figure, with the switc
ID: 2209497 • Letter: A
Question
A.An LR circuit is hooked up to a battery as shown in the figure, with the switch initially open. The resistance in the circuit is R=160Ohm, the inductance is L=3.10H, and the battery maintains a voltage of E=33.0V. At time t=0 the switch is closed.
B. What is the current through the circuit after the switch has been closed for t=1.47?10-2s?
C. What is the voltage across the inductor after the switch has been closed for t= 1.47?10-2 seconds?
A.An LR circuit is hooked up to a battery as shown in the figure, with the switch initially open. The resistance in the circuit is R=160Ohm, the inductance is L=3.10H, and the battery maintains a voltage of E=33.0V. At time t=0 the switch is closed. B. What is the current through the circuit after the switch has been closed for t=1.47?10-2s? C. What is the voltage across the inductor after the switch has been closed for t= 1.47?10-2 seconds?Explanation / Answer
In the question we are given R=100O, L=3.30H, E=36.0V, t=43.2ms The inductor & resistor voltages are given by: v.L(t) = Eo·exp(-tR/L) = 9.7V v.R(t) = Eo(1-exp(-tR/L)) = 26.3V So RL circuit current is: i(t) = v.R(t)/R = 0.26A And resistor power is: Pr(t) = (v.R(t))²/R = 6.9W Inductor energy stored is given by: E.L(t) = ½L·i(t)² = 0.1J Energy dissipated in resistor is: Er(t) = t·Pr(t) = 0.3J Battery work is energy stored & dissipated: W(t) = Er(t) + E.L(t) = 0.4J
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