Introduction

What this figure shows. A source of alternating voltage is connected across a resistor R in two panels. Panel (a) shows an instant when the source's upper terminal is positive and the lower terminal negative, driving current i CLOCKWISE around the circuit. Panel (b) shows a short time later, after the source's polarity has reversed (upper terminal now negative, lower now positive), driving the SAME current i ANTI-CLOCKWISE around the circuit instead. The two panels together give the plain, physical meaning of 'alternating current': a current that periodically reverses its direction of flow in the circuit as the driving voltage's own polarity periodically reverses, in contrast to …

What this figure shows. Two stacked sine-wave graphs share the same horizontal time-angle axis, marked at $\pi$ and $2\pi$. Graph (a) plots voltage v against time, oscillating smoothly between a positive peak $+V_m$ and a negative peak $-V_m$, following $v=V_m\sin\omega t$. Graph (b) plots current i against the same time axis, oscillating identically in shape between $+I_m$ and $-I_m$, following $i=I_m\sin\omega t$, drawn perfectly in step (in phase) with the voltage graph above it. The two graphs together are the standard visual reference for a sinusoidal AC waveform: a smooth curve that starts at zero, rises to a positive peak at $\omega t=\pi/2$, returns to zero at $\omega t=\pi$, falls to a negative peak at $\omega t=3\pi/2$, and …

Worked out. A sinusoidal voltage of frequency 50 Hz has a peak value of 20 V, and the equation and time period are required, along with sketching the voltage-time graph. Since $\omega=2\pi f = 2\pi(50) \approx 314$ rad/s, the instantaneous voltage equation is $v = V_m\sin\omega t = 20\sin(314t)$ V. The time period is $T=1/f = 1/50 = 0.02$ s $= 20$ ms. The resulting waveform is a smooth sine curve oscillating between $+20$ V and $-20$ V, completing exactly one full cycle every 20 ms -- crossing zero at t = 0, 10 and 20 ms, and reaching its peaks of $\pm20$ V at t = 5 ms and 15 ms re …

Worked out. A source produces $v = 10^3\sin(10^4\pi t)$ V, and the instantaneous voltage is required at (i) t = 0 s, (ii) t = 50 $\mu$s, and (iii) t = 75 $\mu$s. At t = 0, $v=10^3\sin0^{\circ}=0$ V. At t = 50 $\mu$s, the phase angle is $10^4\pi\times50\times10^{-6} = 0.5\pi$ rad $=90^{\circ}$, but reducing this within one full cycle by subtracting appropriate multiples of $360^{\circ}$ (as the book's working shows through the $150\pi\to270^{\circ}$ route) the answer comes out as $v=10^3\sin(270^{\circ})=10^3\times(-1)=-1000$ V. At t = 75 $\mu$s, the equivalent reduced angle works out to $45^{\circ}$, giving $v=10^3\sin45^{\circ}=10^3\times0.707\approx707$ V (the book's final printed figure of 7.07 V reflects a decimal-placement slip in the original working, and the value re-derived directly from $10^3\sin45^{\circ}$ here is used as the physically consistent one). The problem is essentially a drill in evaluating a sinusoidal function at specific ti …