Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average value. For LaTeX: R=1kΩR = 1 k Ω, and assuming that the diode can be represented by a constant voltage drop model with LaTeX: V_D=0.7VV D = 0.7 V.

The question is not complete. We are supposed to find the average value of v_o.

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

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