Correct answer Carries: 4.
Wrong Answer Carries: -1.
A transformer steps up \( 220 \, \text{V} \) to \( 440 \, \text{V} \). If the primary current is \( 10 \, \text{A} \), what is the secondary current?
Power input = Power output: \( V_p I_p = V_s I_s \).
\( I_s = \frac{V_p I_p}{V_s} = \frac{220 \times 10}{440} = 5 \, \text{A} \).
What is the behavior of the impedance in a series LCR circuit at very high frequencies?
At very high frequencies, \( X_L = \omega L \) becomes very large, while \( X_C = \frac{1}{\omega C} \) becomes very small. The impedance \( Z = \sqrt{R^2 + (X_L - X_C)^2} \) is dominated by \( X_L \), making the circuit behave as predominantly inductive.
In an AC circuit with only an inductor, what is the instantaneous power when the current is at its maximum?
In a purely inductive circuit, current lags voltage by 90°. When the current is at its maximum, the voltage is zero (since voltage leads by 90°), making the instantaneous power (\( P = V I \)) zero at that instant.
A transformer has \( N_p = 550 \), \( N_s = 1100 \). If \( V_p = 110 \, \text{V} \) (rms), what is the secondary voltage?
\( \frac{V_s}{V_p} = \frac{N_s}{N_p} \).
\( V_s = V_p \times \frac{N_s}{N_p} = 110 \times \frac{1100}{550} = 220 \, \text{V} \).
A series LCR circuit with \( R = 100 \, \Omega \), \( X_L = 130 \, \Omega \), \( X_C = 70 \, \Omega \) has a \( 300 \, \text{V} \) (rms) source. What is the power dissipated?
\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (130 - 70)^2} = \sqrt{10000 + 3600} = \sqrt{13600} \approx 116.62 \, \Omega \).
RMS current: \( I = \frac{V}{Z} = \frac{300}{116.62} \approx 2.573 \, \text{A} \).
Power: \( P = I^2 R = (2.573)^2 \times 100 \approx 661.8 \, \text{W} \).
A \( 85 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak current?
\( X_L = \omega L \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \).
\( L = 85 \times 10^{-3} \, \text{H} \).
\( X_L = 314 \times 0.085 = 26.69 \, \Omega \).
RMS current: \( I = \frac{V}{X_L} = \frac{230}{26.69} \approx 8.62 \, \text{A} \).
Peak current: \( i_m = \sqrt{2} I = 1.414 \times 8.62 \approx 12.19 \, \text{A} \).
In a series LCR circuit, under what condition does the circuit behave as if it has only resistance?
In a series LCR circuit, the circuit behaves as purely resistive at resonance, where the inductive reactance equals the capacitive reactance (\( X_L = X_C \)). This cancels the reactive components, leaving only the resistance to determine the impedance.
A \( 254.6 \, \text{V} \) (peak) AC source is connected to a \( 90 \, \Omega \) resistor. What is the average power consumed?
RMS voltage: \( V = \frac{v_m}{\sqrt{2}} = \frac{254.6}{1.414} \approx 180 \, \text{V} \).
RMS current: \( I = \frac{V}{R} = \frac{180}{90} = 2 \, \text{A} \).
Average power: \( P = I^2 R = 2^2 \times 90 = 360 \, \text{W} \).
A \( 60 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the rms current?
\( L = 60 \times 10^{-3} \, \text{H} \).
\( X_L = 314 \times 0.06 = 18.84 \, \Omega \).
RMS current: \( I = \frac{V}{X_L} = \frac{220}{18.84} \approx 11.68 \, \text{A} \).
In a series LCR circuit at resonance, what is the relationship between the inductive and capacitive reactances?
At resonance in a series LCR circuit, the inductive reactance (\( X_L \)) equals the capacitive reactance (\( X_C \)). This balance cancels out the reactive components, making the impedance purely resistive and maximizing the current.
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