The circuit shown has been operating for a long time. The instant after the switch in the circuit labeled S is opened, what is the voltage across the inductor V_(L) and which labeled point (A to B) of the inductor is at a higher potential? Take R_(1)=4.0Omega, R_(2)=8.0Omega, and L=2.5H.

The circuit shown has been operating for a long time. The instant afte

1.	The circuit shown has been operating for a long time. The instant after the switch in the circuit labeled S is opened, what is the voltage across the inductor V_(L) and which labeled point (A to B) of the inductor is at a higher potential? Take R_(1)=4.0Omega, R_(2)=8.0Omega, and L=2.5H.
Answer» `V_(L)=12 V`, POINT A is at the higher POTENTIAL `V_(L)=12 V`, Point B is at the higher potential `V_(L)=6 V`, Point A is at the higher potential `V_(L)=6 V`, Point B is at the higher potential Solution :When switch is closed for long TIME inductor will acts as short circuit `I=E/(R_(1)\|\|R_(2))=8/3 Amp`. Just after switch is opened. Current in indouctor cannot change instantaneously applying loop law `12-V_(L)-IR=0` `12-V_(L)-4xx8/3=0 implies V_(L)=6` volt so their is `6V` rise ACROSS inductor.

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The circuit shown has been operating for a long time. The instant after the switch in the circuit labeled S is opened, what is the voltage across the inductor V_(L) and which labeled point (A to B) of the inductor is at a higher potential? Take R_(1)=4.0Omega, R_(2)=8.0Omega, and L=2.5H.

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