question_answer

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 or B) of the inductor is at a higher potential? Take, \[{{R}_{1}}=4.0\Omega ,{{R}_{2}}=8.0\Omega \] and \[L=2.5H\].

A)  \[{{V}_{L}}=12V,\] point A is at the higher potential

B)  \[{{V}_{L}}=12V,\] point B is at the higher potential

C)  \[{{V}_{L}}=6V,\] point A is at the higher potential

D)  \[{{V}_{L}}=12V,\] point B is at the higher potential

Correct Answer: D

Solution :

 Current in the inductor before opening4 S' \[I=\frac{12}{\left( \frac{4\times 8}{4+8} \right)}=\frac{9}{2}=4.5A\] Since current in inductor does not change instantly, therefore, just after opening ?S?  \[12+{{V}_{L}}-I{{R}_{1}}=0\] \[12+{{V}_{L}}-(4.5)\,(4)=0\] \[{{V}_{L}}=6V\]with 'B; at higher potential.