Correct answer Carries: 4.
Wrong Answer Carries: -1.
A \( 12 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the capacitive reactance?
\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \).
\( C = 12 \times 10^{-6} \, \text{F} \).
\( X_C = \frac{1}{314 \times 12 \times 10^{-6}} \approx 265.3 \, \Omega \).
In a step-up transformer, how does the current in the secondary coil compare to the primary coil, assuming ideal conditions?
In an ideal step-up transformer (where \( V_s > V_p \) and \( N_s > N_p \)), power is conserved (\( V_p I_p = V_s I_s \)). Since the secondary voltage is higher, the secondary current must be lower than the primary current (\( I_s = I_p \times \frac{N_p}{N_s} \)), where \( \frac{N_p}{N_s} < 1 \).
A \( 55 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the rms current?
\( X_L = \omega L \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \).
\( L = 55 \times 10^{-3} \, \text{H} \).
\( X_L = 314 \times 0.055 = 17.27 \, \Omega \).
RMS current: \( I = \frac{V}{X_L} = \frac{230}{17.27} \approx 13.32 \, \text{A} \).
A \( 150 \, \Omega \) resistor is connected to a \( 300 \, \text{V} \) (peak) AC supply. What is the rms current in the circuit?
Peak voltage: \( v_m = 300 \, \text{V} \).
RMS voltage: \( V = \frac{v_m}{\sqrt{2}} = \frac{300}{1.414} \approx 212.1 \, \text{V} \).
RMS current: \( I = \frac{V}{R} = \frac{212.1}{150} \approx 1.414 \, \text{A} \).
A \( 160 \, \text{V} \) (rms) AC source supplies a \( 80 \, \Omega \) resistor. What is the average power consumed?
RMS current: \( I = \frac{V}{R} = \frac{160}{80} = 2 \, \text{A} \).
Average power: \( P = I^2 R = 2^2 \times 80 = 320 \, \text{W} \).
A \( 120 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the inductive reactance?
\( X_L = \omega L \), \( \omega = 2\pi f \).
\( f = 50 \, \text{Hz} \), \( L = 120 \times 10^{-3} \, \text{H} \).
\( \omega = 2 \times 3.14 \times 50 = 314 \, \text{rad/s} \).
\( X_L = 314 \times 0.12 = 37.68 \, \Omega \).
A series LCR circuit has \( L = 3.5 \, \text{H} \), \( C = 8 \, \mu\text{F} \). What is the resonant angular frequency?
\( \omega_0 = \frac{1}{\sqrt{L C}} \).
\( L = 3.5 \, \text{H} \), \( C = 8 \times 10^{-6} \, \text{F} \).
\( \omega_0 = \frac{1}{\sqrt{3.5 \times 8 \times 10^{-6}}} = \frac{1}{\sqrt{28 \times 10^{-6}}} \approx 188.98 \, \text{rad/s} \).
In a purely capacitive AC circuit, the current leads the voltage by what phase angle?
For a capacitor, \( i = i_m \sin (\omega t + \frac{\pi}{2}) \), \( v = v_m \sin \omega t \).
Phase difference: \( \phi = \frac{\pi}{2} \), so current leads voltage by \( 90^\circ \).
A series LCR circuit has \( R = 40 \, \Omega \), \( L = 5 \, \text{H} \), \( C = 80 \, \mu\text{F} \). What is the resonant frequency in Hz?
\( L = 5 \, \text{H} \), \( C = 80 \times 10^{-6} \, \text{F} \).
\( \omega_0 = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-4}}} = 50 \, \text{rad/s} \).
\( f_0 = \frac{50}{2 \times 3.14} \approx 7.96 \, \text{Hz} \).
In an AC circuit with a series combination of resistor, inductor, and capacitor, what is the condition for the circuit to be purely resistive?
For an LCR series circuit to be purely resistive, the net reactance must be zero (\( X_L - X_C = 0 \)), which occurs at resonance when \( X_L = X_C \). This makes the impedance equal to the resistance (\( Z = R \)), and the circuit behaves as if only resistance is present.
Are you sure you want to submit your answers?