P 5.4-12 The circuit shown in Figure P 5.4-12 contains an adjustable resistor. The resistance R can be set to any value in the range 0 ≤ R ≤ 100 kΩ.

(a) Determine the maximum value of the current ia that can be obtained by adjusting R. Determine the corresponding value of R.

(b) Determine the maximum value of the voltage va that can be obtained by adjusting R. Determine the corresponding value of R.

(c) Determine the maximum value of the power supplied to the adjustable resistor that can be obtained by adjusting R. Determine the corresponding value of R.

 

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Figure P 5.4-12

 

First simplify circuit by adding 12 kΩ and 24 kΩ resistors in series producing a 36 kΩ resistor and then adding the 18 kΩ and the 36 kΩ in parallel

 

 

So the circuit now looks like

Now do a source transformation and we get

Finally, we can add the two power sources and we have the simpliest circuit

We can now write down what ia and va are.  From Ohm’s law

 

And from voltage division

 

(a) Determine the maximum value of the current ia that can be obtained by adjusting R. Determine the corresponding value of R.

 

To maximize , denominator needs to be as small as possible.  So set R to 0.

 

 

(b) Determine the maximum value of the voltage va that can be obtained by adjusting R. Determine the corresponding value of R.

 

 

To maximize , let’s take first derivative with respect to R and set equal to zero.

 

 

 

This will be zero if R approaches infinity, then , which in turn allows

.  So maximum  results when R is a maximum, so  R is 100 kΩ

 

 

(c) Determine the maximum value of the power supplied to the adjustable resistor that can be obtained by adjusting R. Determine the corresponding value of R.

 

Power to R can be found by

 

Again take first derivative and set equal to zero

 

 

 

 

This equals zero when

So the power then is

 

 

 

 

 

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This page last updated on March 7, 2019