• # question_answer The potential difference between points A and B in the circuit shown in the figure will be: A) 1 V                              B) 2 V                  C)      3V                  D)      4 V

From kirchoffs second law, it follows that: $I=\frac{10-5}{25+15+2.5+5+2.5}=\frac{5}{50}=0.1\,A$ Potential difference across A and B                      $=0.1\times (25+15)=4V$ Hence, the correction option is (d).