P5.5-1 The part of the circuit shown in Figure P5.3-1a to the left of the terminals can be reduced to its Norton equivalent circuit using source transformations and equivalent resistance. The resulting Norton equivalent circuit, shown in Figure P5.3-1b, will be characterized by the parameters: and

a) Determine the values of and .

b) Given that , determine the maximum values of the voltage, v, and of the power, p = vi.

Answers: , , max v = 10 V and max p = 1.25 W

Figure P 5.5-1

Let’s first do a source transformation on the 0.25 A source and the 50 Ω resistor and we get the circuit

Combine the two 50 Ω resistors and the two voltage sources and we get

Now do another source transformation and we get

Add the two resistors in parallel

So

And the circuit is

a) Determine the values of and .

Now we are given : and , by comparing the short circuit current and the Thevenin resistance we have

And

From first equation we can find

From second equation

b) Given that , determine the maximum values of the voltage, v, and of the power, p = vi.

So the final circuit looks like

v is either or So we can find the currents by current division

So v is

So, v will be maximum when denominator is smallest which will occur when R₂ is largest, so let R₂ go to ∞

Max power occurs when , so and power is

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This page last updated on March 8, 2019