question_answer

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

A)  4 V                                        

B)  8 V

C)  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\[=\frac{24}{3+3}=4A\] Let potential of point \[O\] shown in figure is \[{{V}_{0}}\] then using Ohms law                 \[{{V}_{0}}-{{V}_{a}}=3\times 4=12\,\,V\]                            ... (i) Now, current through \[5\Omega \] resistor\[=\frac{24}{5+1}\]                                                                 \[=4A\] So           \[{{V}_{0}}-{{V}_{b}}=4\times 1=4V\]                     ... (ii) From Eqs. (i) and (ii), we get                 \[{{V}_{b}}-{{V}_{a}}=12-4=8\,\,V\]