question about parallel dc powersources

Current will flow out of the higher-voltage source into the lower-voltage source.

This won't happen if each voltage source happens to have a diode right at its output. In that case, no current will flow, but if there is a load, the higher-voltage source will try to deliver all the current.

Voltage drops across the resistors will follow Ohm's Law. If you pull enough current out of the 65-volt source, you'll bring its voltage below 61 volts and then the 61-volt source will get to deliver some current into the load also.

Hi, Rob. Thanks for being up front about this being a homework question. Your honesty deserves a helpful response.

fixed font or M$ Notepad):

| ___ A | .--|___|-o----. | +| 2.5 | | | --- | | | 65V- | | | | | .-. | .----' | | |20 | | ___ | | | | | .--|___|-' '-' | | +| 2.0 | | | --- | | | 61V- | | | | B | | '----o--------o----' | (created by AACircuit v1.28.6 beta 04/19/05

In order to use the tools you've learned so far, you'd need to replace the two voltage sources with two internal resistances (everything to the left of A and B) with one voltage source and one series resistance. Then you'll have a circuit like this:

| | ___ A | .--|___|--o------. | | Rs | | V| .-. | --- 20| | | - | | | | '-' | | B | | '---------o------' | | (created by AACircuit v1.28.6 beta 04/19/05

There's a method for replacing tehse multiple voltage sources with one equivalent voltage source for circuit analysis. It's called Thevenin's Theorem.

This link will show you exactly how to solve your problem. It's easy once you know how to do it.

Good luck in your studies Chris

"Rob L."

** The outcome in the real world depends on the characteristics of the actual DC voltage sources.

Are you dealing with "ideal" voltage sources or practical ones ?

** Must know if the sources are "ideal" ones.

** OK:

Assuming "ideal" sources ( as usual for math problems), first ignore the

20 ohm load and find the open cct voltage for the combined sources. This will be 61 volts PLUS the 4 volts difference divided by the ratio of 2.5:2.

Then find the short cct current of the combined sources, this is simply

65/2.5 plus 61/2 (because the short prevents current flowing between the two sources).

The open cct voltage divided by the short cct current gives the internal resistance of the combined source.

The rest is now simple.

....... Phil

parallel

if

2.5

question

A single nodal equation will solve the circuit. The node is where the three resistors connect, i.e.the top end of the 20 ohm resistor. The sum of the currents flowing in this node must equal zero. Therfore the current from 65V source must be (65 -Vn)/2.5. where Vn is the voltage on the node. Similarly the current from the 61V source must be (61 - Vn)/2 and the sum of those two currents must be equal to the current flowing into the 20 ohm resistor, Vn/20. I.E., (65 - Vn)/2.5 + (61 -Vn)/2 = Vn/20. Solving for the voltage on the node, Vn = 59.474Volts. Its trivial to figure what each current is once the node voltage is found. In this case, both voltage sources contribute to the the load current.

Consider the case where the load resistor is large or disconnected. Now, the higher voltage source will feed the lower one. Without a load, a current of

0.888Amps would flow between them draining the higher source. That's why you don't want to parallel voltage sources. Consider what whould happen if the source resistances were 0.001ohms as might be the case with high power sources. Bad News! Bob

You need to draw the 'equivalent circuit' using voltage sources and resistances. Like Chris explained. Then you can analyse it.

And yes, the higher voltage source will indeed supply some current into the lower voltage one.

Graham

"Poopie Bear Fuckwit "

** No it does not.

Both sources deliver current into the load.

2.2 amps from the 65 volt source ans 0.75 amps from the 61 volt one.

......... Phil

Kirchoff's laws and a little applied algebra will get you the answer.

Bye. Jasen