................... Consider the circuit shown in the figure. The battery has em

Question

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Consider the circuit shown in the figure. The battery has emf epsilon = 53 volts and negligible internal resistance. The inductance is L = 0.8 H and the resistances are R 1 = 12 Ohm and R2 = 9.0. Initially the switch S is open and no currents flow. Then the switch is closed. What is the current in the resistor R 1 just after the switch is closed? Express your answer using two significant figures. Part B After leaving the switch closed for a very long time, it is opened again. Just after it is opened, what is the current in R 1? Express your answer using two significant figures.

Explanation / Answer

initially the inductor will act as an open circuit, so the currect throught R1 = 75 /12 = 25/4A

but aftera long time it will act as no inductance component, so current = 75 / (12*9/(12+9)) = 7*75/36 = 525 / 36A

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