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Understand the Problem
The question paper is related to Electrical Engineering, specifically focusing on electrical circuits. It contains several parts asking for calculations and theoretical knowledge regarding R-L-C circuits, power factor, selectivity, impedance, and circuit properties.
Answer
The power factor is approximately $0.707$.
Answer for screen readers
The power factor is approximately $0.707$.
Steps to Solve
- Calculate the Impedance of the R-L-C Circuit
To find the impedance ($Z$) of a series R-L-C circuit, use the formula:
$$ Z = \sqrt{R^2 + (X_L - X_C)^2} $$
Where:
- $R = 10 , \Omega$ (resistance)
- $X_L = 20 , \Omega$ (inductive reactance)
- $X_C = 10 , \Omega$ (capacitive reactance)
Substituting the values:
$$ Z = \sqrt{10^2 + (20 - 10)^2} $$ $$ Z = \sqrt{100 + 10^2} $$ $$ Z = \sqrt{100 + 100} $$ $$ Z = \sqrt{200} $$ $$ Z = 10\sqrt{2} , \Omega $$
- Calculate the Power Factor
The power factor ($PF$) is defined as:
$$ PF = \frac{R}{Z} $$
Using earlier results:
$$ PF = \frac{10}{10\sqrt{2}} $$ $$ PF = \frac{1}{\sqrt{2}} $$ $$ PF = 0.707 $$
- Conclusion on Power Factor
The power factor of the series R-L-C circuit is thus:
Result: The power factor is approximately 0.707.
The power factor is approximately $0.707$.
More Information
The power factor indicates the efficiency of the circuit in converting electric power into useful work output. A power factor of 0.707 suggests that the circuit is moderately efficient in using power.
Tips
- Confusing reactance values: Ensure you correctly differentiate between inductive and capacitive reactance.
- Neglecting phase angles: Remember to consider how resistance, inductance, and capacitance affect the phase of the current and voltage.
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