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.You need to login to perform this action.
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