A) zero
B) 2 V
C) 3.5 V
D) 4.5 V
Correct Answer: D
Solution :
\[2I+2{{I}_{1}}-6=0\] \[I+{{I}_{1}}=3~~\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( 1 \right)\]\[\left( I-{{I}_{1}} \right)1+6-2{{I}_{1}}=0\] \[I-3{{I}_{1}}=-6~~~~~~~~~~~~....\left( 2 \right)\] From (1) & (2) \[\therefore \,\,\,{{I}_{1}}=9/4\] \[{{V}_{A}}-{{V}_{B}}=IR\] \[=\,\,\,9/4\times 2\] \[=9/2\,\,=\,\,4.5\]You need to login to perform this action.
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