A sinusoidal source of peak voltage and frequency feeds a resistor of constant…

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electronics, Instrumentation & Communication · Semiconductor devices and analogue circuits

Electricity, Magnetism and Electromagnetism, Ch Alternating current: peak and rms values, the oscilloscope and rectification

Problem

A sinusoidal source of peak voltage

V_0 = 17.0\;\mathrm{V}

and frequency

50\;\mathrm{Hz}

feeds a resistor of constant resistance

R = 20\;\Omega

through a single ideal diode (no voltage across it when it conducts, no current through it when it blocks). The voltage across the resistor is therefore

V(t) = V_0\sin\omega t

for

0 \le t < T/2

and

V(t) = 0

for

T/2 \le t < T

, repeating with the period

T

of the source, where

\omega = 2\pi/T

. ((a)) By averaging

V^2

over one full period, using

\sin^2\theta = \tfrac12(1-\cos 2\theta)

, show that the root-mean-square voltage of this half-wave output is

V_0/2

. ((b)) Evaluate

V_{\mathrm{rms}}

for the numbers given. ((c)) Find the mean power dissipated in the resistor. ((d)) Find the mean value

\langle V\rangle

of the half-wave output, and say in one sentence why it is not equal to

V_{\mathrm{rms}}

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