Refer to Figure P2.27 and assume that vS = 15 V and RS = 100 ?. For iT = 0, 10, 20, 30, 80, and 100 mA:

Question

a. Find the total power supplied by the ideal source.

b. Find the power dissipated within the non-ideal source.

c. How much power is supplied to the load resistor?

d. Plot the terminal voltage vT and power supplied to the load resistor as a function of terminal current iT .







What will be an ideal response?

Answer 1

Known quantities:

vs=15V, Rs=100 Ohms, iT= 0, 10, 20, 30, 80, 100 mA.

The circuit in Figure P2.27.

Find:

a) The total power supplied by the ideal source

b) The power dissipated within the non-ideal source

c) How much power is supplied to the load resistor

d) Plot vT and power supplied to R0 as a function of iT.

Analysis:

a) The power supplied by the ideal source is equal to the current through the loop times the 15V of the supply. From current lowest to highest the power supplied would be:

0W 0.15W 0.3W 0.45W 1.2W 1.5W

b) The power dissipated within the non-ideal source is the power dissipated by Rs which can be found using P=i2*r. From current lowest to highest the power dissipated would be:

0W 0.01W 0.04W 0.09W 0.64W 1W

c) The power supplied to the load resistor is equal to the total power supplied minus the power dissipated by the non-ideal source. From current lowest to highest the power supplied would be:

0W 0.14W 0.26W 0.36W 0.56W 0.5W

d) For the vT plot Ohm’s Law can be used to find the voltage drop across Rs which is vT = 15 – 100iT. For the power plot, the data from part c can be used directly.