NEETPhysicsElectromagnetic Induction
A branch XY of an electrical circuit carries a current of 2 A flowing from X to Y. The branch contains a 3 resistor, a 4 V battery (with its positive terminal facing X), and an inductor of 2 H connected in series. If the potential difference V_X - V_Y is 14 V , what is the rate of change of current in the branch and its nature?
Options
- AIncreasing at 6 A/s
- BIncreasing at 2 A/s
- CDecreasing at 2 A/s
- DDecreasing at 6 A/s
Correct answer
B. Increasing at 2 A/s
Step-by-step solution
Apply Kirchhoff's Voltage Law (KVL) from point X to point Y. The current i flows from X to Y, so the potential drops across the resistor by iR . The battery's positive terminal faces X, so traversing from X to Y goes from positive to negative, giving a potential drop of E . The potential drop across the inductor is L di dt . The KVL equation is: V_X - iR - E - L di dt = V_Y Rearranging for V_X - V_Y : V_X - V_Y = iR + E + L di dt Substitute the given values ( V_X - V_Y = 14 V , i = 2 A , R = 3 , E = 4 V , L = 2 H )