Correct answer Carries: 4.
Wrong Answer Carries: -1.
A \( 50 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the peak current?
\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \).
\( C = 50 \times 10^{-6} \, \text{F} \).
\( X_C = \frac{1}{314 \times 50 \times 10^{-6}} \approx 63.7 \, \Omega \).
RMS current: \( I = \frac{V}{X_C} = \frac{220}{63.7} \approx 3.45 \, \text{A} \).
Peak current: \( i_m = \sqrt{2} I = 1.414 \times 3.45 \approx 4.88 \, \text{A} \).
A series LCR circuit has \( R = 20 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 60 \, \Omega \). What is the phase angle?
\( \tan \phi = \frac{X_C - X_L}{R} = \frac{60 - 40}{20} = 1 \).
\( \phi = \tan^{-1}(1) = 45^\circ \).
Since \( X_C > X_L \), current leads voltage.
A series LCR circuit has \( R = 50 \, \Omega \), \( X_L = 80 \, \Omega \), \( X_C = 60 \, \Omega \). What is the power factor?
\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + (80 - 60)^2} = \sqrt{2500 + 400} = \sqrt{2900} \approx 53.85 \, \Omega \).
Power factor: \( \cos \phi = \frac{R}{Z} = \frac{50}{53.85} \approx 0.928 \).
A \( 20 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the peak current?
\( C = 20 \times 10^{-6} \, \text{F} \).
\( X_C = \frac{1}{314 \times 20 \times 10^{-6}} \approx 159.2 \, \Omega \).
RMS current: \( I = \frac{V}{X_C} = \frac{230}{159.2} \approx 1.445 \, \text{A} \).
Peak current: \( i_m = \sqrt{2} I = 1.414 \times 1.445 \approx 2.04 \, \text{A} \).
What is the role of resonance in the operation of a radio receiver?
In a radio receiver, resonance in an LCR circuit is used to tune to a specific frequency. By adjusting the capacitance or inductance, the circuit’s resonant frequency matches the desired station’s signal, amplifying it while filtering out others.
A series LCR circuit has \( L = 5 \, \text{H} \), \( C = 20 \, \mu\text{F} \). What is the resonant angular frequency?
\( \omega_0 = \frac{1}{\sqrt{L C}} \).
\( L = 5 \, \text{H} \), \( C = 20 \times 10^{-6} \, \text{F} \).
\( \omega_0 = \frac{1}{\sqrt{5 \times 20 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-4}}} = 100 \, \text{rad/s} \).
A \( 180 \, \text{V} \) (rms) source supplies a \( 90 \, \Omega \) resistor. What is the peak current?
RMS current: \( I = \frac{V}{R} = \frac{180}{90} = 2 \, \text{A} \).
Peak current: \( i_m = \sqrt{2} I = 1.414 \times 2 \approx 2.828 \, \text{A} \).
Why does a purely capacitive AC circuit not dissipate power despite having current flow?
In a purely capacitive circuit, the current leads the voltage by 90°. The instantaneous power oscillates, but the average power over a cycle is zero because the energy is stored during one half-cycle and returned during the other, with no net dissipation.
A series LCR circuit with \( R = 50 \, \Omega \) is at resonance with a \( 230 \, \text{V} \) (rms) source. What is the rms current?
At resonance, \( Z = R = 50 \, \Omega \).
RMS current: \( I = \frac{V}{R} = \frac{230}{50} = 4.6 \, \text{A} \).
In an AC circuit with only a capacitor, what is the nature of the current when the voltage is at its peak?
In a purely capacitive circuit, the current leads the voltage by 90°. When the voltage is at its peak (maximum positive or negative), the current is zero because it reaches its peak 90° earlier and crosses zero at the voltage peak.
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