AC Voltage Applied to a Resistor

What this figure shows. A simple series circuit diagram showing a resistor of resistance R connected across the two terminals of an AC source (drawn as a circle with a sine-wave symbol inside, or the standard AC-source circle-with-tilde symbol), with a single loop of connecting wire joining them, and the instantaneous emf e and instantaneous current i both labelled on the connecting wires -- the simplest possible AC circuit, containing only R and nothing else (no L, no C), used to derive the purely-resis …

What this figure shows. Two sinusoidal curves plotted on the SAME set of axes (e and i both on the vertical axis, $\omega t$ on the horizontal axis, spanning one full cycle from 0 to $2\pi$): one curve for the instantaneous emf e (peak $e_0$) and one for the instantaneous current i (peak $i_0$), drawn as two sine waves of possibly different amplitude but starting together at the origin, rising together, peaking together at $\omega t=\pi/2$, crossing zero together at $\omega t=\pi$, reaching their minima together at $\omega t=3\pi/2$, and returning to zero together at $\omega t=2\pi$ -- visually demonstrating that e and i reach zero, their maximum, and their minimum values SIMULTANEOUSLY at every point in the cycle, i.e. there is no horizontal (pha …

What this figure shows. A phasor diagram with a fixed horizontal reference axis OX, showing TWO phasor arrows drawn EXACTLY along the same line/direction from the origin O, both making the same angle $\omega t$ with OX: one arrow representing the peak emf $e_0$ and the other representing the peak current $i_0$, drawn overlapping or immediately alongside each other (never separated by any angle) to emphasise that the phase angle between the current and voltage phasors through a pure resistor is exactly zero -- unlike the inductor and capacitor cases that follow, where the two phasors will be …

Worked out. An alternating voltage $e=140\sin(3142t)$ is connected across a pure resistor of $R=50\,\Omega$. Comparing with $e=e_0\sin\omega t$ gives $\omega=3142$ rad/s and $e_0=140$ V. The frequency follows from $\omega=2\pi f$: $f=3142/(2\times3.142)=500$ Hz. The rms voltage is $e_{rms}=e_0/\sqrt2=140/1.414\approx98.99$ V, and since a resistor is purely Ohmic for both AC and DC, the rms current follows directly from Ohm's law applied to rms values: $i_{rms}=e_{rms}/R=98.99/50\approx1.98$ A. This example demonstrates that rms voltage and rms current obey the ordinary Ohm's law relation for a resistor, exactly as …