Download PDF by Noel M. Morris (auth.): Electrical Circuits and Systems

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Q. 14 is the equivalent output circuit of a transistor amplifier. Determine (a) the value of the load resistor RL to give maximum power transfer, and (b) the magnitude of the power delivered into the load. Solution The output resistance of the network is obtained by measuring the resistance between points A and B, the current generator being meanwhile replaced by its internal resistance (infinity, since it is a current generator) and the load resistor being disconnected. In the case considered, the equivalent output resistance is 10 X 90 r=--krl=9krl 10 + 90 For maximum power transfer to occur, then (b) The current in R L is h r 9 = 10 x --rnA= 10 x - = 5 rnA RL + r 9 +9 42 Electrical Grcuits and Systems and the power consumed by R L is PL =(5 X 10- 3 ?

F. f. f. acting in loop 3. 27 all have positive signs since they act in a direction to produce a clockwise circulating current within their own mesh. R 1 1 = self-resistance of loop 1 = R 1 + R 2 + R 3 R 2 2 =self-resistance of loop 2 = R 2 + R 4 + R 5 R 3 3 =self-resistance ofloop 3 = R 3 + R 5 + R 6 In order to determine the self-resistance of each loop, replace each generator in the loop by its internal resistance, and open-circuit all other loops, then calculate the loop resistance. Circuit Theorems 51 R12 = R 2 1 = (-1) x the mutual resistance associated with loops 1 and 2 = -- R 2 R 1 3 = R 3 1 = ( -1) x the mutual resistance associated with loops 1 and 3 = - R 3 R 2 3 = R 3 2 = ( -1) x the mutual resistance associated with loops 2 and 3 =- R 5 The mutual resistance associated with two loops is the resistance which causes a potential drop in one loop due to a current in the other.

2 Solution First, we assign currents I 1 , I 2 , and I 3 to the three branches joining at node C, noting that / 3 =/ 1 +/2 . f. and resistor, the direction in which the arrow points indicating which end of the resistor has the more positive potential if the current flows in the direction we have selected. The circuit equations can now be formed. 43 A Since I 1 and I 3 have positive signs associated with them, it follows that these currents flow in the directions shown on the diagram. Current I 2 has a negative sign associated with it, so that it flows in the opposite direction to that chosen, that is, it flows from node C to node D.

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