A branch of an electrical circuit between points M and N contains an ideal inductor of 1 H , a battery of 3 V , and a resistor of 2 connected in series. The current flows from M to N. The negative terminal of the battery is connected towards M and the positive terminal towards N. The current i (in A) varies with time t (in s) as a straight line passing through the coordinates (t=0, i=1) and (t=2, i=5) on an i-t graph

Step-by-step solution

From the given coordinates (0, 1) and (2, 5) , the rate of change of current is the slope of the i-t graph: di dt = 5 - 1 2 - 0 = 2 A/s The current at t = 1 s can be found from the linear relation i = 1 + 2t : i = 1 + 2(1) = 3 A Applying Kirchhoff's Voltage Law from M to N: V_M - L di dt + E - iR = V_N Note that the battery is traversed from the negative to the positive terminal, so its potential change is +3 V . Substituting the values: V_M - (1)(2) + 3 - (3)(2) = V_N V_M - 2 + 3 - 6 = V_N V_M - 5 = V_N V_M - V_N