A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100 sqrt2 sin (1000 t). The power factor of the combination is ______.

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A resistor of 500 Ω and inductance of 0.5 H is in series with an AC source which is given by V = 100 sqrt2 sin (1000 t). The power factor of the combination is 1/sqrt2. Explanation: Substitute the given values from L = 0.5 H .....(1) R = 500 Ω .....(2) V = 100sqrt(2) sin(1000t) .....(3) On comparing with, V = V_0 sin(wt) We get, w = 1000s^-1 .....(4) Put the values in equation (1) and (4) and calculate: Let, Inductive reactance XL XL = w_L = 1000 $\times$ 0.5 XL = 500 Ω ......(5) Since, Power factor cos phi `cos phi = R/sqrt(X_L^2 + R^2) ....(7) $\therefore$ cos phi = 500/sqrt(500^2 + 500^2) cos phi = 500/(500sqrt(2) cos phi = 1/sqrt(2) Therefore, P.F = cos phi = 1/sqrt(2)`

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