Introduction to Alternating Current (AC)

AC voltages can be easily and efficiently converted from one voltage to another using transformers.

True

<p>v = vm sin ωt</p> Signup and view all the answers

<p>Nicola Tesla</p> Signup and view all the answers

<p>vm sin ωt</p> Signup and view all the answers

<p>Nicola Tesla</p> Signup and view all the answers

<p>i = (vm/R) sin ωt</p> Signup and view all the answers

<p>The total impedance (Z) is calculated using the formula $Z = R^2 + (X_L - X_C)^2$.</p> Signup and view all the answers

<p>A negative phase difference indicates that the current lags behind the voltage in the circuit.</p> Signup and view all the answers

<p>Power is calculated using the formula $P = I^2 R$, which results in $P = (40A)^2 imes 3Ω = 4800W$.</p> Signup and view all the answers

<p>Capacitive reactance (X_C) is calculated using the formula $X_C = \frac{1}{2 \pi u C}$, showing its inverse relationship with frequency.</p> Signup and view all the answers

<p>The power factor is defined as cos(φ), where φ is the phase angle; hence, a phase angle of -53.1° results in a power factor of 0.6.</p> Signup and view all the answers

<p>The power input must equal the power output, meaning there are no energy losses.</p> Signup and view all the answers

<p>If the secondary coil has more turns than the primary, the voltage is stepped up.</p> Signup and view all the answers

<p>The current decreases in the secondary coil compared to the primary coil.</p> Signup and view all the answers

<p>In a step-down transformer, the voltage is reduced (Vs &lt; Vp) while the current increases (Is &gt; Ip).</p> Signup and view all the answers

<p>Equation (7.35) represents the relationship between primary and secondary current and voltage ratios.</p> Signup and view all the answers

<p>The assumptions include small primary resistance, equal flux linkage, and small secondary current.</p> Signup and view all the answers

<p>A step-up transformer has more turns in the secondary coil than in the primary, increasing voltage.</p> Signup and view all the answers

<p>High efficiency (over 95%) allows transformers to be used effectively in power systems with minimal energy loss.</p> Signup and view all the answers

<p>The current reaches its maximum value one-fourth of a period later than the voltage.</p> Signup and view all the answers

<p>No, an inductor does not consume power, as the average power over a complete cycle is zero.</p> Signup and view all the answers

<p>The inductive reactance is 7.85 Ω.</p> Signup and view all the answers

<p>The rms current is 28 A.</p> Signup and view all the answers

<p>Current flows briefly to charge the capacitor, then stops as the voltage across it builds up.</p> Signup and view all the answers

<p>Current continuously flows as the capacitor charges and discharges with the alternating voltage.</p> Signup and view all the answers

<p>The instantaneous power is represented by $p_L = -\frac{i_m v_m}{2} \sin(2\omega t)$.</p> Signup and view all the answers

<p>The average power is zero, indicating that the inductor does not consume energy over a complete cycle.</p> Signup and view all the answers

<p>212 Ω</p> Signup and view all the answers

<p>1.04 A</p> Signup and view all the answers

<p>The peak current is $\sqrt{2}$ times the rms current.</p> Signup and view all the answers

<p>Capacitive reactance is halved and current is doubled.</p> Signup and view all the answers

<p>The glow of the light bulb decreases.</p> Signup and view all the answers

<p>The iron rod magnetizes, increasing the magnetic field and the inductance.</p> Signup and view all the answers

<p>Higher capacitive reactance results in lower current, dimming the lamp.</p> Signup and view all the answers

<p>Higher reactance leads to lower current flow.</p> Signup and view all the answers

<p>The current falls to zero.</p> Signup and view all the answers

<p>It limits or regulates the current but does not completely prevent the flow of charge.</p> Signup and view all the answers

<p>The instantaneous voltage is given by the formula $v = \frac{q}{C}$.</p> Signup and view all the answers

<p>The equation is $v_m \sin(\omega t) = \frac{q}{C}$.</p> Signup and view all the answers

<p>The current is given by $i = \omega C v_m \cos(\omega t)$.</p> Signup and view all the answers

<p>It is expressed as $i_m = \omega C v_m$.</p> Signup and view all the answers

<p>It plays the role of resistance.</p> Signup and view all the answers

<p>The alternative expression is $i = i_m \sin(\omega t + \frac{\pi}{2})$.</p> Signup and view all the answers