Question

In the circuit shown in Fig. the battery provides an E = 15 V


In the circuit shown in Fig. the battery provides an E = 15 V, the inductance is L = 0.5 H, and the resistances are R1 = 12Ω and R2 = 9Ω. Initially the switch S is open, and At t = 0, suddenly the switch is closed. At that moment what is the current going through R1 in A? 

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Answer #1

At t= 0 , inductor will behave as open circuit i.e. no current passes through it...

Current through R1 = E/R = 15/12 = 1.25 A

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