In the circuit shown charge q varies with time t as q = t^2 - 5 , where q is in coulomb and t is in second. At time t = 3 second, voltage V_ AB in volt will be

Step-by-step solution

Given charge on the capacitor, q = t^2 - 5 . The current i in the circuit is the rate of flow of charge: i = dq dt = d dt (t^2 - 5) = 2t The rate of change of current is: di dt = d dt (2t) = 2 A/s At time t = 3 s : Charge, q = (3)^2 - 5 = 4 C Current, i = 2(3) = 6 A Rate of change of current, di dt = 2 A/s Applying Kirchhoff's Voltage Law from point A to point B: V_A - V_C - V_L - V_R = V_B V_A - q C - L di dt - iR = V_B V_A - V_B = q C + L di dt + iR Substituting the given values ( C = 4 F , L = 0.5 H , R = 2 ): V