By M. B.; Ellis, J. Pamela Ellis
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5(a)). We need not enquire how the driving network develops the current, suffice to say that it is available at its terminals. 5(b)) and adjusted until the current is reduced to zero. The driving network is de livering no current and it is therefore operating as if on open circuit. The voltage at its terminals, v , is the open-circuit voltage. The voltage across the external resistor VR = iRb = 0 since i = 0. 5(c)) without affecting either of them. 1) ext making clear that the external source has been set to the open-circuit voltage of the driving network.
The correct procedure in such cases may be seen by con sidering an example. 3(a) we wish to find vn. We shall sup press the two independent sources in turn, one at a time. 3(b). 5 A and so 6i\ is 3 V. A further formal application of the principle of superposition could now be made but it is fairly evident that we have a 3 V source opposing a 5 V source with the resultant being 2 V. 3(c), the left most resistor is in parallel with the short circuit and so i\ is zero. It follows that the dependent source is dead.
6 V across the two resistors in parallel. 2 The duality of other quantities has also emerged. A short circuit needs to be trans lated to an open circuit, a series connection to a parallel connection and a node to a mesh. From what we already know of the behaviour of capacitors and inductors we can see that they are duals also: After translating voltage to current and vice versa the equations become the same if inductance is translated to capacitance and vice versa. By exchanging current for voltage, short circuit for open circuit, and so on, Norton's theorem is produced from Thevenin's theorem and so we see that the theorems themselves are duals.