How Do Current Sources and Resistors Interact in a Circuit?

  • Thread starter Thread starter SmallDot
  • Start date Start date
  • Tags Tags
    Current Sources
AI Thread Summary
In the discussion, participants explore how to combine current sources in parallel and their effects on resistors in a circuit. They clarify that current sources can be summed to create an equivalent source, while resistors also combine in a similar manner. The impact of a 36A current source on a 20-ohm resistor is examined, emphasizing that current flows through paths of lower resistance. The voltage drop across resistors in parallel is equal, and Kirchhoff's Current Law (KCL) is applied to analyze current flow at nodes. Overall, understanding these interactions is crucial for analyzing circuit behavior effectively.
SmallDot
Messages
4
Reaction score
0

Homework Statement


Combining the current sources of the given circuit. I've attached the circuit diagram below.
circuit 1.PNG

The Attempt at a Solution


[/B]
Since the sources are in parallel I can just add up the values to get an equivalent source and same for the resistors. I'm curious in the proof of this fact. With two current sources and one resistor, I can see how the sources can be combined. But for cases like this, how does the 36A affect the 20 ohms resistor? How do the currents circulate in the circuit?
 
Physics news on Phys.org
SmallDot said:

Homework Statement


Combining the current sources of the given circuit. I've attached the circuit diagram below.
View attachment 102827

The Attempt at a Solution


[/B]
Since the sources are in parallel I can just add up the values to get an equivalent source and same for the resistors. I'm curious in the proof of this fact. With two current sources and one resistor, I can see how the sources can be combined. But for cases like this, how does the 36A affect the 20 ohms resistor? How do the currents circulate in the circuit?
Current will want to flow across the path of less resistance. So a resistor with high resistance will have less current going through it, while there is more current flow across a resistor with less resistance. The only time current will not flow through a resistor is if there is a shorted wire in parallel with the circuit.
Imagine driving on a 3-lane road. The first lane on the left side is congested, the middle lane has 1-2 cars moving quickly, and the rightmost side is closed due to construction..The movement of the cars can be analogous to current flow in a circuit, and of course, we'd want to take the lane that is moving and isn't congested..
The 20Ω and 5Ω will have the same voltage drop across them because they're connected in parallel. Each current sources will have current flowing across both resistors until their voltages are same.
Formula to keep in mind: V=I*R
Hope this somewhat helps :-p
 
Last edited:
SmallDot said:
But for cases like this, how does the 36A affect the 20 ohms resistor? How do the currents circulate in the circuit?
The upper horizontal line is one single node, so to make the arrangement clearer you could redraw the schematic showing that node as a single point, with 5 wires going to it. Likewise for the lower horizontal line.

Current doesn't accumulate at a node; KCL says that at any node, current in = current out.

The 6A source pumps 6A into the upper node, the 36A source pumps a further 36A in, while the 12A source pumps 12A out of that upper node. Applying KCL to the upper node tells you how much current must, therefore, be leaving that node via other paths, i.e., through the resistors.

you could just as correctly call it a 12A current sink
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
Back
Top