In an electric circuit, an inductor of inductance 0.5 H and resistance 50\(\Omega\) is connected to an alternating current source of frequency 60 Hz. Calcul...

The impedance of the circuit can be calculated using the formula: Z = √(R^2 + X^2) Where R is the resistance of the inductor and X is the reactance of the inductor, which is given by: X = 2πfL Where f is the frequency of the AC source and L is the inductance of the inductor. Substituting the given values, we get: X = 2π(60)(0.5) = 188.5 Ω R = 50 Ω Z = √(50^2 + 188.5^2) = 195.0 Ω Therefore, the impedance of the circuit is 195.0 Ω. To explain it in simpler terms, the impedance of the circuit is like the total resistance that the AC source "sees" when it is connected to the circuit. It takes into account both the resistance and the reactance (which is like a resistance but specific to AC circuits) of the circuit. In this case, the inductor has a reactance of 188.5 Ω which, when combined with the resistance of 50 Ω, gives an impedance of 195.0 Ω.