Consider the circuit shown in the figure. http://session.masteringphysics.com/pr

Question

Consider the circuit shown in the figure. http://session.masteringphysics.com/problemAsset/1543243/1/p24.46.jpg The battery has emf ? = 89 volts and negligible internal resistance. The inductance is L = 0.1 H and the resistances are R 1 = 12 ? and R 2 = 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

V = 89 V A.) As soon as the switch is closed, the inductor acts as a open circuit, Thus we have , Reqv. = R1 = 12 ( I am unable to see the unit of resistance, so I suppose it's in ohm) Thus , Current = V/R = 89/12 = 7.42 A , (if resistance is in Kohm, then answer is 7.42 mA) B.) as the switch is closed for long time, the inductor acts as a short circuit, Now as the switch is opened, we have an RL circuit , Thus , Potential drop across L = 89 V , Thus current through R1 = 89/(R1+R2)= (89)/( 12+9) = 4.24 A

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