By A. M. P. Brookes and P. Hammond (Auth.)

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**Extra resources for Advanced Electric Circuits**

**Example text**

A current VjR is driven through R. ©t f 5 Volts 1 = 1 Amp^- F I G . 5 Combine these results. d. Facross them; a current / flows out of the current source, and a current {VjR)-I out of the voltage source. The powers delivered are therefore IxV and [(VjR)-I]xV from the current and the voltage sources respectively. When these are equal, R =2-5 Q. ■■*-£2 x 1 = 2-5Q ■-GH In the circuit in Fig. 6 it is required to find the current in the 1 Q resistor. A B 3ft ;w F I G . 6 The following four methods could be used: (a) Thevenin's theorem (b) Nodal analysis (two unknowns) c NETWORK THEOREMS 29 (c) Superposition (d) Mesh analysis (four unknowns) By method (a) branch AB, shown separately in Fig.

It is required to find the power delivered by each source and the value of load resistance for which each source would deliver the same amount of power. By principle of superposition the current in R is the sum of the current which flows when voltage source alone acts, and of the current which flows when the current source alone acts. In each case the other, dead, generator must be left connected. When / only acts all of its current must flow through the dead voltage generator because it is a short-circuit.

1 (a) Express the voltage excitation in complex form. (b) Set up the circuit equation. 32 ADVANCED ELECTRIC CIRCUITS (c) Obtain the modulus and phase angle of /, and then turn the resulting complex quantity back into a cosine. ^/is:(a)3e^, [V(3)-j]e^, ,3-V(3)+j (b) / = QW x' l+j (c) modulus 1-14, phase angle -6-7° ( =38-3° -45°) M4cos(ft>/-6-7°)A 2. 2 shows two current generators feeding a load. Carry out steps similar to (a), (b) and (c) in Question 1, for the node potential. ) 3 cosurt 2cosM-30°) Volts Volts FIG.