Understanding Current & Voltage in Circuits A & B w/ Diode and Battery

[KuriousKid]

So I'm trying to understand the relationship between current & voltage as well as the basic electricity laws and Diode operation. Attached are 2 circuit diagrams for reference. Assume that Diodes can withstand against whatever voltage/current the circuit generates.

Can Anyone explain me what will happen in each circuit A & B?

Circuit A : What would be the value of Current and voltage across the load of 1 ohm?
Circuit B : What would be the value of Current and voltage across the load of 1 ohm?

 

[vk6kro]

The first circuit is too difficult to give a result for. When you put different voltage batteries in parallel, the larger voltage one will try to charge the other. So, don't do that.

The second one is also too difficult.
The current in the 1 ohm resistor would be about 5 amps (6 volts minus 1 volt for the diode divided by 1 ohm) but this would exceed the rating of the 6 volt battery.

In each case, you need to know the internal resistance of the batteries to work out what would really happen.

Just to get some sort of an answer, I assumed all the batteries had an internal resistance of 0.2 ohms.
This assumes the diodes have a drop of 1 volt at this current:

[PLAIN]http://dl.dropbox.com/u/4222062/currents.PNG

 

[KuriousKid]

I was under impression that if V*I = same for both sides it will behave as symmetrical circuit and will add up current in 1 ohm resistor.

Here I don't understand why we can't get current addition in 1 ohm resistor per Ohm's/Kirchoff's laws?

What arrangement should be made in the circuit so we can get the sum of currents in 1 ohm resistor and approximately 5 volt across it, so that we can get approximately VI + VI = 2 VI kind of thing?

I don't want to go for series connection of batteries.

 

[vk6kro]

The current figures next to each battery are some sort of maximum current rating. That current will only flow if the load is correct for the voltage. That is the only way you can "tell" a battery how much current to deliver.

The problem is that you have two batteries in parallel with no resistance in the circuit between them. (This only refers to the outer loop.)

There is a net voltage of 3 volts (ie 6 minus 3) but zero resistance so the current will be infinite unless we know something about the internal resistance of the batteries.

There may be some current in the 1 ohm resistor, but until we know what the internal resistances of the batteries are, we can't calculate the current.

In the second circuit, the diode stops any current flowing into the 3 volt battery, but the current taken from the 6 volt battery is about 5 amps which exceeds the current rating of the 6 volt battery.

 

[KuriousKid]

Thank you for your reply.
Now considering your circuit diagram if R2 & R3 = 6 Ohm and R1 = 10 Ohm. What would be the current flowing through R1 and voltage across it?

The first circuit is too difficult to give a result for. When you put different voltage batteries in parallel, the larger voltage one will try to charge the other. So, don't do that.

The second one is also too difficult.
The current in the 1 ohm resistor would be about 5 amps (6 volts minus 1 volt for the diode divided by 1 ohm) but this would exceed the rating of the 6 volt battery.

In each case, you need to know the internal resistance of the batteries to work out what would really happen.

Just to get some sort of an answer, I assumed all the batteries had an internal resistance of 0.2 ohms.
This assumes the diodes have a drop of 1 volt at this current:

[PLAIN]http://dl.dropbox.com/u/4222062/currents.PNG[/QUOTE]

 

[vk6kro]

I guess you mean the circuit without the diodes.

The current from the 6 volt battery is 423.077 mA.

This splits into two currents.

346.154 mA goes into the 10 ohm resistor and 76.923 mA goes into the 3 volt battery via the 6 ohm resistor.

The voltage across the 10 ohm resistor is 3.46 volts.

 

[KuriousKid]

No, I mean to say the Circuit with Diodes in place. Won't the current from both batteries will flow into 10 Ohm resistor [addition of current?] and then again split exactly same way as it merged and flow back into Batteries?

I guess you mean the circuit without the diodes.

The current from the 6 volt battery is 423.077 mA.

This splits into two currents.

346.154 mA goes into the 10 ohm resistor and 76.923 mA goes into the 3 volt battery via the 6 ohm resistor.

The voltage across the 10 ohm resistor is 3.46 volts.

 

[vk6kro]

Assume 1 volt drop across the diode in the left loop.

The current will be (6 volts minus 1 volt) divided by (6 ohms plus 10 ohms). so, 5 / 16 or 0.312 amps.

The voltage across the 10 ohms resistor is then 0.312 amps times 10 ohms or 3.12 volts. (V = I * R)

So, the diode on the right is reverse biased and cannot conduct any current.

 

[KuriousKid]

That's what confuses me. I thought if both sources have same power (P=V*I=6x3=3x6=18) they might be considered as identical sources. Doesn't it work based on Power?

Assume 1 volt drop across the diode in the left loop.

The current will be (6 volts minus 1 volt) divided by (6 ohms plus 10 ohms). so, 5 / 16 or 0.312 amps.

The voltage across the 10 ohms resistor is then 0.312 amps times 10 ohms or 3.12 volts. (V = I * R)

So, the diode on the right is reverse biased and cannot conduct any current.

 

[vk6kro]

That's what confuses me. I thought if both sources have same power (P=V*I=6x3=3x6=18) they might be considered as identical sources. Doesn't it work based on Power?

No, it doesn't work that way. The battery produces a voltage, but it can't control the current. The current is a result of what load you put on the battery.

Batteries have effectively got an internal resistance which you can use to find out how the battery will behave under load.

This internal resistance is effectively in series with the battery, so that even if you short circuit the battery the current will not be infinite.

With an alkaline AA cell, for example, if you short circuit it, the current will be about 8 amps, so you can work out that this battery has an internal resistance of 0.1875 ohms.

It is because the internal resistance has 1.5 volts across it and 8 amps flowing in it so it must have a resistance of 1.5 volts / 8 amps or 0.1875 ohms. This is Ohm's Law.

This is pretty good, but you can see why a bunch of AA cells probably couldn't start a car. They might be able to supply 12 volts, but their internal resistance is too high to produce the 100 amps or more it takes to turn the starter motor in a car.

 

[KuriousKid]

I see. from your answer got another question. What does AC generator produces? Current or Voltage?

No, it doesn't work that way. The battery produces a voltage, but it can't control the current. The current is a result of what load you put on the battery.

 

[vk6kro]

I see. from your answer got another question. What does AC generator produces? Current or Voltage?

It produces both, but the current doesn't flow unless there is a path for it to flow in.

So, as you might expect, if you have wires coming out of the generator and they are not connected to anything, there is no current flowing.

There would be an AC voltage on the wires and you could measure this (in volts) with a suitable meter.

If you did connect a resistor across the output wires, there would be an AC current flowing.