question_answer

In the circuit shown in figure, switch \[{{S}_{1}}\] is initially closed and \[{{S}_{2}}\] is open. Find\[{{V}_{a}}-{{V}_{b}}\].

A) 4 V                              

B) 8 V             

C) (e) 12 V                        

D) 16 V            

Correct Answer: B

Solution :

Switch \[{{S}_{2}}\] is open so capacitor is not in circuit Current through \[3\,\Omega \] resistor \[=\,\,\,\frac{24}{3+3}\,\,4\,A\] Let potential of point ?O? shown in the figure is      \[{{V}_{0}}\] then using Ohm?s law                   \[{{V}_{a}}-{{V}_{a}}=3\times 4=12\,\,V\]                   ?. (i) Now current through \[5\,\Omega \] resistor \[=\,\,\frac{24}{5+1}\,\,=\,\,4\,A\] So \[{{V}_{a}}-{{V}_{b}}=4\times 1\,\,=\,\,4\,V\]                      ?. (ii) From equations (i) and (ii)               \[{{V}_{b}}-{{V}_{a}}\,\,=\,\,12-4\,\,=\,\,8\,\,V\]