For full points/best answer you must provide me with a detailed explicit solutio
ID: 1383170 • Letter: F
Question
For full points/best answer you must provide me with a detailed explicit solution. Please define all variables and equations. Thank you
Consider the circuit sketched above. The battery has emf epsilon = 25 volts and negligible internal resistance. The inductance is L = 0.80 H and the resistances are R1 = 12 ohm and R2 = 9.0ohm. Use Figure 30.5 to answer the following questions: Initially the switch S is open and no currents flow. Then the switch is closed. What is the current in the resistor R1 just after the switch is closed? After leaving the switch closed for a long time, it is opened again. Just after it is opened, what is the current in R1?Explanation / Answer
initially the inductor is not charged so acts as broken wire
hence current in resistot r1 = 25/12 = 2.0833 amp
after long time the inductor gets charged fully and acts like a wire
current through r1 just after reopened = E/r1+r2 = 25/12+9 = 1.1904 amp
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