In the circuit of Figure 10.54, the voltage source vs(t) varies with time as shown by the waveform of Figure 10.55. Compute, sketch, and express vLOAD(t) as a sum of unit step functions for 0 ≤ t ≤ 5 s

Question

In the circuit of Figure 10.54, the voltage source [latex]v_{s}(t) [/latex] varies with time as shown by the waveform of Figure 10.55. Compute, sketch, and express [latex]v_{LOAD}(t) [/latex] as a sum of unit step functions for 0 ≤ t ≤ 5 s

Accepted Answer

We replace the given circuit shown below with its Thevenin equivalent.

For the Thevenin equivalent voltage at different time intervals is as shown below.

[latex] v_{TH}(t)=\begin{cases}\frac{12}{18}v_{s}(t) = \frac{2}{3} imes 60 = 40 V & 0 4 s \end{cases} [/latex]

and

[latex] v_{LOAD}(t) = \frac{10}{20}v_{TH}(t) = 0.5 v_{TH}(t) = \begin{cases} 0.5 × 40 = 20 V & 0 4 s \end{cases} [/latex]

The waveform of the voltage across the load is as shown below.

The waveform above can now be expressed as a sum of unit step functions as follows:

[latex] v_{LOAD}(t) = 20u_{0}(t) - 20u_{0}(t - 1 ) + 40u_{0}(t - 1) - 40u_{0}(t - 1) - 40u_{0}(t - 3) + 20u_{0}(t - 4) -20u_{0}(t - 3) + 20 u_{0}(t - 4) + 20 u_{0}(t - 4) [/latex]

[latex] = 20u_{0}(t) + 20u_{0}(t - 1) - 60 u_{0}(t - 3) + 40 u_{0}(t - 4) [/latex]

Question 10.p.1: In the circuit of Figure 10.54, the voltage source vs(t) var...

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