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I have a problem dealing with equivalent circuit. I don't know why when the resistors that is parallel with the voltage source can be ignore as well as one in series with current source. Of course the voltage in the load is no difference but what about the power. Having a resistor parallel with voltage source brings a difference in total resistance compared with having none resistor parallel with voltage source

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For example. If I place a resistor \$R_1\$ to 2 terminal a-b the thing will turn out for the voltage source having no parallel resistor

$$I_1 = \frac{V_s}{R_1+R}$$

but for one having parallel resistor, it will be

$$I_1 = \frac{R_p}{R_1+R+R_p}(\frac{1}{R_1}+\frac{1}{R_p}+\frac{1}{R})V_s$$

So although the voltage between a-b terminal is the same but the current is not. So how can it be considered equivalent as well as in the current source case. Please give me some proper explaination

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  • \$\begingroup\$ Can you tell us the source of your pictures, maybe we need the context to understand this well. \$\endgroup\$ – diverger Nov 3 '14 at 15:25
  • \$\begingroup\$ What is the series resistance of an ideal voltage source? Now what is the combined resistance of that and Rp in parallel? These two will get you closer to the answer. \$\endgroup\$ – Brian Drummond Nov 3 '14 at 15:53
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In your equivalent resistance equations, you are missing the fact that a voltage source has impedance of 0, and a current source has infinite impedance. Anything in parallel with 0 impedance still results in 0 impedance. Likewise, anything in series with infinite impedance still results in infinite impedance.

To help convince yourself, try putting some real number on your top example. Pick some values for the voltage, Rp, and R. Now put different loads on the result and compute what the voltage on the load is with and without Rp. You will see that Rp doesn't matter. The voltage across the voltage source is the specified voltage, by definition of what a voltage source is. Put another way, a ideal 10 V source always gives you 10 V, whether someone else is drawing 10 mA, 10 A, or nothing from the same voltage source.

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  • \$\begingroup\$ if it's so. Why can the power delivered to load be the same. I mean, the inner resistance will be difference when the source is connected to load. So the share of load's power and the power dissipated by inner resistor will be different but the power the source supplies is unchanged. \$\endgroup\$ – aukxn Nov 3 '14 at 15:34
  • \$\begingroup\$ The OP mentioned the power, so i think he maybe confuse with the word "equivalent", here "equivalent" means for the external circuit, right? Because apparently, the loss with and without \$R_{p}\$ is different. \$\endgroup\$ – diverger Nov 3 '14 at 15:34
  • \$\begingroup\$ @diverger: Yes, the power produced by the source will be different, but this is not apparent to someone performing measurements on the external terminals of the black box. \$\endgroup\$ – Olin Lathrop Nov 3 '14 at 15:42
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The current and the power delivered by the voltage source will not be the same with the parallel resistor, however the voltages at the nodes (and the currents beyond the resistor/source pair) will be calculated correctly.

You run into a similar issue with Thevenin equivalents- they yield the correct answers as to the voltage at the terminals, but the internal power dissipated will not be the same.

You just have to decide not to look too closely inside the 'box'.

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  • \$\begingroup\$ So using source transformation is correct when we concern only about the load's voltage but not load's current or power? I am talking about the external circuit behavior here. According to the class I attend. The behavior of the internal circuit is different but the external will be the same, including the power, meaning that the power delivered to the load shoud also be the same \$\endgroup\$ – aukxn Nov 3 '14 at 15:44
  • \$\begingroup\$ The load voltage and power will indeed be identical. However, if you look at the source- an ideal current source will zero dissipation when shorted- whilst an ideal voltage source will have zero dissipation when open. \$\endgroup\$ – Spehro Pefhany Nov 3 '14 at 15:56
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I think( for voltage source) parallel resistor can be ignored because that resistor has same voltage as voltage source.So if there is any possibility to flow current through this then total current of the circuit needs to flow through it as it needs to maintain voltage difference same to voltage source.Then current doesn't have any other directions to flow so it's acted as series connection of resistor which is not acceptable.That's why current doesn't chose resistor's path so it will act as open circuit which can be ignored. (For current source) series resistor can be ignored because current flows through that resistor is just equal to the current source's supply and if other elements in that source's loop gain same current then series resistor can't drop any voltage,if it's drop any voltage then current wouldn't be same as source for other other elements.That's why series resistor can't drop any voltage as current source needs to maintain contact current flow.So series resistor will work as short circuit which can be ignored in series connection.

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