In the circuit shown, when the current ' i ' is 3 A and increasing at the rate of 1 A/s the measurement of the potential difference between A and B is 12 V . But when the same current 3 A is decreasing at the rate of 1 A/s , the measured potential difference V_ AB between A and B is 6 V . The value of 'R' in the circuit is

Step-by-step solution

The potential difference between points A and B can be written using Kirchhoff's Voltage Law as: V_A - iR - L di dt = V_B V_ AB = V_A - V_B = iR + L di dt Case 1: When the current is 3 A and increasing at 1 A/s , we have i = 3 A and di dt = 1 A/s . Given V_ AB = 12 V , substituting the values: 12 = 3R + L(1) 3R + L = 12 ... (i) Case 2: When the current is 3 A and decreasing at 1 A/s , we have i = 3 A and di dt = -1 A/s . Given V_ AB = 6 V , substituting the values: 6 = 3R + L(-1) 3R - L = 6 ... (ii) Adding equation