To determine the power factor of the circuit consisting of a resistor and an inductor connected in series, we first need to find the total impedance of the circuit and then calculate the power factor.
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Given:
- Resistance (R) = 25 Ω (resistor)
- Inductive Reactance (X_L) = ? (due to the coil)
- Resistance of the coil (r) = 50 Ω (but this is in series with the resistor)
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Total resistance (R_total): \[ R_{\text{total}} = R + r = 25, \Omega + 50, \Omega = 75, \Omega \]
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Calculating the Inductive Reactance (X_L): The inductive reactance \(X_L\) can be calculated using the formula: \[ X_L = 2 \pi f L \] However, we need to know the frequency \(f\) to calculate it. For now, let's proceed theoretically.
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Total Impedance (Z): Since the resistor and inductor are in series, the total impedance \(Z\) is calculated as: \[ Z = \sqrt{R_{\text{total}}^2 + X_L^2} \]
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Calculate Power Factor (PF): The power factor (PF) is defined as the cosine of the phase angle \(\phi\) between the voltage and current, given by: \[ PF = \cos(\phi) = \frac{R_{\text{total}}}{Z} \]
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Substituting the Values: We can express the Power Factor as: \[ PF = \frac{R_{\text{total}}}{\sqrt{R_{\text{total}}^2 + X_L^2}} \]
However, without the frequency \(f\), we can't find \(X_L\) directly. For illustrative purposes, if we assume some frequency, say 50 Hz (which is common):
- Calculate \(X_L\) for \(L = 150 \text{ mH} = 0.150 \text{ H}\): \[ X_L = 2 \pi (50, \text{Hz}) (0.150, \text{H}) = 2 \pi (50)(0.150) \approx 47.1, \Omega \]
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Calculating Impedance: \[ Z \approx \sqrt{(75)^2 + (47.1)^2} \] \[ Z \approx \sqrt{5625 + 2216.41} \approx \sqrt{7841.41} \approx 88.6, \Omega \]
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Final Power Factor Calculation: \[ PF = \frac{R_{\text{total}}}{Z} = \frac{75}{88.6} \approx 0.847 \]
So, the power factor of the circuit is approximately 0.847, which indicates a lagging power factor due to the presence of the inductor.
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