We need to determine how the power factor of an \( L, R \) circuit changes when the frequency of the AC source is altered. Let's go through the problem step by step.
The initial AC source is given by \( E = 25 \sin 1000 \, t \, \text{V} \) with a power factor of \( \frac{1}{\sqrt{2}} \). We first need to establish some key concepts:
Initially, the circuit's angular frequency is:
\(\omega_1 = 1000 \, \text{rad/s}\)
When the source is changed to \( E = 20 \sin 2000 \, t \, \text{V} \), the new angular frequency is:
\(\omega_2 = 2000 \, \text{rad/s}\)
Now, let's analyze what happens to the power factor:
Thus, the new power factor of the circuit when the frequency is increased to 2000 rad/s is \( \frac{1}{\sqrt{5}} \).
Step 1: Determine Initial Reactance \( X_L \): Since the initial power factor \( \cos \theta = \frac{1}{\sqrt{2}} \), we have \( \tan \theta = 1 \), meaning \( X_L = R \). With the initial angular frequency \( \omega_1 = 1000 \, \text{rad/s} \), we can write \( X_L = \omega_1 L = R \).
Step 2: Calculate Reactance at New Frequency: For the new frequency, \( \omega_2 = 2000 \, \text{rad/s} \), the new inductive reactance becomes:
\[ X'_L = \omega_2 L = 2 \omega_1 L = 2R \]
Step 3: Determine New Power Factor: With the new reactance \( X'_L = 2R \), we find \( \tan \theta' = \frac{X'_L}{R} = 2 \), which gives:
\[ \cos \theta' = \frac{1}{\sqrt{1 + (2)^2}} = \frac{1}{\sqrt{5}} \]
Conclusion: The new power factor is therefore:
\[ \frac{1}{\sqrt{5}} \]



Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.