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(C) From the given circuit diagram,we have two resistors in series,$R_1 = 40\, \Omega$ and $R_2 = 40\, \Omega$.Total resistance $R_{net} = R_1 + R_2 =

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$0.8$

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(C) From the given circuit diagram,we have two resistors in series,$R_1 = 40\, \Omega$ and $R_2 = 40\, \Omega$.Total resistance $R_{net} = R_1 + R_2 = 40 + 40 = 80\, \Omega$.The inductive reactance is $X_L = 100\, \Omega$ and the capacitive reactance is $X_C = 40\, \Omega$.The net reactance $X = |X_L - X_C| = |100 - 40| = 60\, \Omega$.The impedance of the circuit is $Z = \sqrt{R_{net}^2 + X^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\, \Omega$.The power factor is given by $\cos \phi = \frac{R_{net}}{Z} = \frac{80}{100} = 0.8$.

The power factor of the circuit as shown in the figure is:

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