A series \( L, R \) circuit connected with an AC source \( E = (25 \sin 1000 \, t) \, \text{V} \) has a power factor of \( \frac{1}{\sqrt{2}} \). If the source of emf is changed to \( E = (20 \sin 2000 \, t) \, \text{V} \), the new power factor of the circuit will be:
- \( \frac{1}{\sqrt{2}} \)
- \( \frac{1}{\sqrt{3}} \)
- \( \frac{1}{\sqrt{5}} \)
- \( \frac{1}{\sqrt{7}} \)
The Correct Option is C
Approach Solution - 1
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:
- The power factor, \( \cos \phi \), of an \( L, R \) circuit is defined as \( \cos \phi = \frac{R}{Z} \), where \( Z \) is the total impedance of the circuit, calculated as \( Z = \sqrt{R^2 + (X_L)^2} \).
- The inductive reactance \( X_L \) is given by \( X_L = \omega L \), where \( \omega = 2\pi f \) is the angular frequency.
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:
- With the initial frequency \( \omega_1 = 1000 \, \text{rad/s} \), the inductive reactance is \( X_{L1} = 1000L \). The power factor is:
- With the new frequency \( \omega_2 = 2000 \, \text{rad/s} \), the inductive reactance changes to \( X_{L2} = 2000L \). Thus, we need to calculate the new power factor:
- Using the initial power factor equation \( \frac{1}{\sqrt{2}} = \frac{R}{\sqrt{R^2 + (1000L)^2}} \), we can analyze that:
- Substituting the condition \( R^2 = (1000L)^2 \) into the expression for \( \cos \phi_2 \):
Thus, the new power factor of the circuit when the frequency is increased to 2000 rad/s is \( \frac{1}{\sqrt{5}} \).
Approach Solution -2
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:
\[ \frac{1}{\sqrt{5}} \]
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