Question

The potential difference between points A and B in the circuit shown in figure will be

​

Answer 1

the potential difference is the 3v

Answer 2

Answer:

(d) 4 V

Explanation:

From the given figure we can see that all the circuit elements are connected in series and we have to calculate the potential difference between point A and point B in the circuit.

According to KCL theorem we know that current entering at a point will be equal to the current leaving that point.

Hence by applying KCL at point A

    $\dfrac{V_B\ -\ V_A}{25+15}\ =\ \dfrac{V_A-V_B+10-5}{2.5+5+2.5}$

$=>\ \dfrac{V_A\ -\ V_B}{40}\ +\dfrac{V_A-V_B+5}{10}\ =\ 0$

$=>\ \dfrac{V_A\ -\ V_B}{40}\ +\dfrac{4(V_A-V_B)+20}{40}\ =\ 0$

$=>\ 5(V_A-V_B)\ =\ -20$

$=>\ V_A-V_B\ =\ -4$

Hence, the potential difference between point A and B will be 4 V.