Correct answer Carries: 4.
Wrong Answer Carries: -1.
In an AC circuit with a series combination of resistor and capacitor, what happens to the phase difference if the frequency approaches infinity?
In an RC series circuit, \( \phi = \tan^{-1} \left( \frac{X_C}{R} \right) \), where \( X_C = \frac{1}{\omega C} \). As frequency (\( \omega \)) approaches infinity, \( X_C \) approaches zero, making \( \phi \) approach 0°, so the circuit becomes nearly resistive.
A series LCR circuit with \( R = 25 \, \Omega \) is at resonance with a \( 200 \, \text{V} \) (rms) source. What is the rms current?
At resonance, \( Z = R = 25 \, \Omega \).
RMS current: \( I = \frac{V}{R} = \frac{200}{25} = 8 \, \text{A} \).
A \( 18 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the rms current?
\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 60 = 376.8 \, \text{rad/s} \).
\( C = 18 \times 10^{-6} \, \text{F} \).
\( X_C = \frac{1}{376.8 \times 18 \times 10^{-6}} \approx 147.3 \, \Omega \).
RMS current: \( I = \frac{V}{X_C} = \frac{110}{147.3} \approx 0.747 \, \text{A} \).
A series LCR circuit has \( R = 100 \, \Omega \), \( X_L = 80 \, \Omega \), \( X_C = 60 \, \Omega \). What is the impedance?
\( Z = \sqrt{R^2 + (X_L - X_C)^2} \).
\( Z = \sqrt{100^2 + (80 - 60)^2} = \sqrt{10000 + 400} = \sqrt{10400} \approx 102 \, \Omega \).
A \( 30 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the peak current?
\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \).
\( C = 30 \times 10^{-6} \, \text{F} \).
\( X_C = \frac{1}{314 \times 30 \times 10^{-6}} \approx 106.1 \, \Omega \).
RMS current: \( I = \frac{V}{X_C} = \frac{220}{106.1} \approx 2.074 \, \text{A} \).
Peak current: \( i_m = \sqrt{2} I = 1.414 \times 2.074 \approx 2.93 \, \text{A} \).
What is the effect on the power factor of an LCR series circuit when the frequency is decreased below resonance?
Below resonance, \( X_C > X_L \), making the circuit capacitive. The phase angle \( \phi = \tan^{-1} \left( \frac{X_C - X_L}{R} \right) \) is positive, and the power factor \( \cos \phi \) decreases (less than 1) as the circuit becomes less resistive.
A transformer has \( N_p = 100 \) and \( N_s = 200 \). If the primary voltage is \( 220 \, \text{V} \) (rms), what is the secondary voltage?
\( \frac{V_s}{V_p} = \frac{N_s}{N_p} \).
\( V_s = V_p \times \frac{N_s}{N_p} = 220 \times \frac{200}{100} = 440 \, \text{V} \).
A \( 50 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the inductive reactance?
\( X_L = \omega L \), where \( \omega = 2\pi f \).
\( f = 60 \, \text{Hz} \), \( L = 50 \times 10^{-3} \, \text{H} \).
\( \omega = 2 \times 3.14 \times 60 = 376.8 \, \text{rad/s} \).
\( X_L = 376.8 \times 0.05 = 18.84 \, \Omega \).
What is the average power dissipated in a purely capacitive circuit over one complete cycle?
Power: \( p_C = i v = i_m v_m \cos (\omega t) \sin (\omega t) = \frac{i_m v_m}{2} \sin (2\omega t) \).
Average over a cycle: \( P_C = \frac{i_m v_m}{2} \langle \sin (2\omega t) \rangle = 0 \), since \( \langle \sin (2\omega t) \rangle = 0 \).
A \( 80 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the rms current?
\( X_L = \omega L \), \( \omega = 2\pi \times 60 = 376.8 \, \text{rad/s} \).
\( L = 80 \times 10^{-3} \, \text{H} \).
\( X_L = 376.8 \times 0.08 = 30.14 \, \Omega \).
RMS current: \( I = \frac{V}{X_L} = \frac{230}{30.14} \approx 7.63 \, \text{A} \).
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