To analyze the parallel circuit with the given parameters, we will begin by calculating the impedance of each branch, then find the current in each branch, the total current supplied, and the phase angle between the supply voltage and the total current.
Given Data
- Voltage \( V = 100 , \text{V} \)
- Frequency \( f = 50 , \text{Hz} \)
- Resistance \( R = 50 , \Omega \)
- Inductance \( L = 0.5 , \text{H} \)
- Capacitance \( C = 100 , \mu\text{F} = 100 \times 10^{-6} , \text{F} \)
Step 1: Calculate Reactances
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Inductive Reactance \( X_L \): \[ X_L = 2 \pi f L = 2 \pi (50) (0.5) = 157.08 , \Omega \]
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Capacitive Reactance \( X_C \): \[ X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (50) (100 \times 10^{-6})} \approx 31.83 , \Omega \]
Step 2: Calculate the Impedance of Each Branch
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Impedance of the Resistance Branch (purely resistive): \[ Z_R = R = 50 , \Omega \]
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Impedance of the Inductive Branch: \[ Z_L = R + jX_L = 50 + j157.08 , \Omega \] \[ |Z_L| = \sqrt{50^2 + 157.08^2} \approx 165.5 , \Omega \]
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Impedance of the Capacitive Branch: \[ Z_C = R - jX_C = 50 - j31.83 , \Omega \] \[ |Z_C| = \sqrt{50^2 + 31.83^2} \approx 58.5 , \Omega \]
Step 3: Calculate Current in Each Branch
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Current through the Resistance Branch (\( I_R \)): \[ I_R = \frac{V}{Z_R} = \frac{100}{50} = 2 , A \]
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Current through the Inductive Branch (\( I_L \)): \[ I_L = \frac{V}{Z_L} = \frac{100}{165.5} \approx 0.604 , A \] The phase angle for this branch: \[ \phi_L = \tan^{-1}\left(\frac{X_L}{R}\right) = \tan^{-1}\left(\frac{157.08}{50}\right) \approx 72.34^\circ \]
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Current through the Capacitive Branch (\( I_C \)): \[ I_C = \frac{V}{Z_C} = \frac{100}{58.5} \approx 1.71 , A \] The phase angle for this branch: \[ \phi_C = -\tan^{-1}\left(\frac{X_C}{R}\right) = -\tan^{-1}\left(\frac{31.83}{50}\right) \approx -32.6^\circ \]
Step 4: Total Current Calculation
Since they are in parallel: \[ I_{total} = I_R + I_L + I_C \] To sum these currents, we will convert them into rectangular form based on their phase angles.
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Convert to Rectangular Form:
- \( I_R = 2 + j0 \)
- \( I_L = 0.604 (\cos 72.34^\circ + j \sin 72.34^\circ) \approx 0.604 (0.307 + j0.951) \approx 0.185 + j0.574 \)
- \( I_C = 1.71 (\cos(-32.6^\circ) + j \sin(-32.6^\circ)) \approx 1.71 (0.848 - j0.54) \approx 1.45 - j0.92 \)
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Add the currents together: \[ I_{total} = (2 + 0.185 + 1.45) + j(0 + 0.574 - 0.92) = 3.635 + j(-0.346) \] Therefore, \[ |I_{total}| = \sqrt{(3.635)^2 + (-0.346)^2} \approx 3.637 , A \]
Step 5: Calculate the Phase Angle of the Total Current
\[ \phi_{total} = \tan^{-1}\left(\frac{-0.346}{3.635}\right) \approx -5.4^\circ \]
Summary of Results
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Branch Currents:
- \( I_R = 2 , A \) (0 degrees)
- \( I_L \approx 0.604 , A \) (approx 72.34 degrees)
- \( I_C \approx 1.71 , A \) (approx -32.6 degrees)
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Total Supply Current:
- \( I_{total} \approx 3.637 , A \)
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Phase Angle of Total Current:
- \( \phi_{total} \approx -5.4^\circ \)
The calculations shown involve estimation terms and numerical approximations; make sure to confirm with accurate values if needed for critical applications!