question_answer

If in the circuit shown below, the internal resistance of the battery is 1.5 W and VP and VQ are the potentials at P and Q respectively, what is the potential difference between the points P and Q                                       [MP PET 2000]

A)             Zero

B)            4 volts (VP > VQ)

C)            4 volts (VQ > VP)

D)            2.5 volts (VQ > VP)

Correct Answer: D

Solution :

                   \[{{R}_{eq}}=\frac{5}{2}\Omega \] \[i=\frac{20}{\frac{5}{2}+1.5}=5A\] Potential difference between X and P, \[{{V}_{X}}-{{V}_{P}}=\left( \frac{5}{2} \right)\times 3=7.5V\]                  ....(i) \[{{V}_{X}}-{{V}_{Q}}=\frac{5}{2}\times 2=5V\]                              ?(ii)       On solving (i) and (ii) \[{{V}_{P}}-{{V}_{Q}}=-2.5\text{ }volt\text{; }{{V}_{Q}}>{{V}_{P}}\]. Short Trick : \[({{V}_{P}}-{{V}_{Q}})=\frac{i}{2}({{R}_{2}}-{{R}_{1}})\]\[=\frac{5}{2}(2-3)=-\,2.5\] Þ \[{{V}_{Q}}>{{V}_{P}}\]