A coil having an inductance of 0.16 H and a resistance of 12 is connected to an alternating voltage source given by V = 200 2 (100 t) , where V is in volts and t is in seconds. The peak current in the circuit and the average power dissipated by the coil are respectively:

Step-by-step solution

The given alternating voltage is V = 200 2 (100 t) . Comparing this with the standard equation V = V₀ ( t) , we get: Peak voltage, V₀ = 200 2 V Angular frequency, = 100 rad/s The inductive reactance X_L is given by: X_L = L = 100 0.16 = 16 The impedance Z of the coil is: Z = R^2 + X_L^2 = 12^2 + 16^2 = 144 + 256 = 400 = 20 The peak current I₀ in the circuit is: I₀ = V₀ Z = 200 2 20 = 10 2 A The RMS current I_ rms is: I_ rms = I₀ 2 = 10 2 2 = 10 A The average power dissipated by the coil is: P_ avg = I_ rms ^2 R = (