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Current Injection

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Dfektv

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Hello,

I'm new to electronics. I have what seems a silly question to you guys.

With respect to the picture attached, how would I inject, say, another amp through the resistors indicated by the additional power supply, whilst maintaining a voltage rise accross them as though there was no additional power supply (i.e. 4V)?

Would I need to reduce the resistance in this section by 1/3 and apply 4V at the additional supply? Or, must the additional supply be greater than 4V by a value proportional to the increase in current (with respect to the resistance)?

If the resistors in this section must be changed, how will it affect the current supplied from the 12V source?

Could I please get some basic calculated figures for this example?

Thanks in advance
 

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I'll take a wild stab at this.

If the additional power supply is set to 4V, then there would be no change in the resistor voltages or current, and no current would be drawn from the additional supply.
 
Sorry about the lack of clarity of the question. crutschow - would that mean that if I wanted to pass an additional 1 amp through the 2 resistors, the additional supply would have to be 6V?
 
Yes. Good news first: If you do a superposition (or similar circuit check), an additional 6V will drive 3A through those resistors instead of the usual 2A. It's also going to lower the current through the other resistors to 1 ½ Amps because that new voltage is opposing your original 12V source.

My first recommendation is to group the 2 resistors that are attached to your additional voltage source into one 2Ω. Then do the same with the other 4 resistors and group them into a single 4Ω resistor. After that, the simplified circuit will be a 4Ω resistor, a 2Ω resistor and the two voltages.

I'll use these names...
R2 = 2Ω resistor
R4 = 4Ω resistor
I2 = current through 2Ω resistor
I4 = current through 4Ω resisor
Vmain = original voltage supply
Vnew = your additional supply


To find the currents you want, you'll have to redraw the circuit 3 different ways: (1) one where Vnew is replaced by a short circuit, (2) one where the Vmain is replaced by a short circuit, and (3) a final drawing where Vnew and Vmain are normal

Part 1. Calculate the currents that would run through R2 and R4 if the Vnew supply was replaced by a wire. Since doing that also shorts out R2, that resistor's current drops to 0 Amps. The 4Ω resistor gets 3Amps from the 12V battery (12Volts / 4Ohms = 3Amps). Write down the currents...

I2 = 0 Amp
I4 = 3 Amps

Part 2. Calculate the currents as if Vmain was replaced by a short circuit wire. Doing that essentially puts R2 and R4 in parallel. Find their currents is a little tougher than part 1, but if you take it step by step, you'll eventually get...

I2 = 3 Amps
I4 = -1.5 Amps
...I4 is negative because this current would be opposing Vmain in the original diagram.

Part 3. Finally, you combine the results from Parts 1 & 2.
I2 = 0 + 3 = 3 Amps
I4 = 3 - 1.5 = 1.5 Amps


...that was hard to illustrate without diagrams, but hopefully it will provide a guide.
 
Last edited:
Thanks DigiTan.

Applying those values to the circuit in the thumbnail I posted,
would that mean that for the two resistors having 3A through them, 1.5A is supplied from the 12V source, and 1.5A supplied from the 6V source, with the 1.5A supplied from the 12V continuing through to the last resistor?

A bit of a mouthful I apologise.
 
That's correct. You'll have 1.5Amps coming out of both sources. The 4Ω part will carry 1.5Amps, and the 2Ω part will carry 3Amps.
 
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