JEE Main 2020 · 5 September · evening · Single correct · +4 / −1
In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is :
- (a) +2 V
- (b) -2 V
- (c) +1 V
- (d) -1 V
Answer and solution
Answer (c) +1 V
Solution
Physics sir · 4 steps
Get the 2 Ω current from junction C, then walk from A to B adding every rise and drop.
- Junction at C. 1 A comes into C from A, and 2 A leaves C towards F. The missing 1 A must enter C through the 2 Ω resistor, so 1 A flows from D down to C.
- From A to D. In the 1 V cell the longer plate is on the C side, so going from A to C the potential rises by 1 V. Current flows from D to C through 2 Ω, so D is higher than C by 1 × 2 = 2 V. That gives V_D = V_A + 3 V.
- From D to B. In the 2 V cell the longer plate is on the D side, so going from D to B the potential falls by 2 V.
V_B - V_A = (+1) + (+1 × 2) + (-2) = +1 V
- So. Point B is at +1 V with respect to A, which is option (c).
The idea to keep. For the potential between two points, walk any path joining them: a cell gives +ε going short plate to long plate, and a resistor gives -IR when you move along the current.
Answer from the NTA final answer key. Solution by PrepWiser.