9780199339136, Chapter 1, Problem 24P

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Chapter Problem 24P Step-by-step solution Step The signal source IS not ideal, considering the fact that the output voltage changes with the change in the load resistance The signal source can be replaced as source voltage in series with an internal resistance, Draw the equivalent R, Figure Step Write the expression for the output voltage of the circuit Simplify further The output voltage 40 mV for 100 kΩ Therefore, substitute 100 kΩ for and 40 mV for equation (1). 100k (2) Comments (2) Anonymous Can you please break down how this was simplified further to get Vo= Anonymous It's because they said that RI/RI Rs/RI IS equal to (RI+Rs)/RI. So then RI/RI so you get Rs/RI 1 Step of 8 The output voltage 10 mV for 10 kΩ load. substitute 10 for R1 and 10 mV for in equation (1) 10k Substitute (40 for (The source voltage is constant.) 100k 100k R 10k 100k 10k 25k 10k 10k 25k 3 10k 25k 250M 50k the value of the source resistance is Step of 8 Substitute 50 kΩ for R. in equation (2) 100k 50 k = m) = 40 m) 2 =60 mV Therefore, the value of the source voltage is 60 mV Step Replace the load resistance with an open circuit Draw the resultant circuit VTH Figure 2 Step of 8 The voltage obtained across the load resistance terminals, with the load open circuited is the Thevenin's As no current flows through the circuit the voltage across the load terminals is equal to the source VTH Substitute 60 mV for VTH mV Therefore, the Thevenin voltage 60 mV Step of 8 Replace the load resistance by short circuit to determine the short circuit Figure Step of 8 The Norton current Substitute 60 mV for and 50 for R, 50 k Therefore, the value of the Norton current